Reversing the chain rule If
F F F is an antiderivative of
f f f , the chain rule gives
d d x F ( g ( x ) ) = F ′ ( g ( x ) ) g ′ ( x ) = f ( g ( x ) ) g ′ ( x ) . \begin{aligned}
&\frac{d}{dx}\,F\bigl(g(x)\bigr) \\[4pt]
&\qquad = F'\bigl(g(x)\bigr)\,g'(x) \\[4pt]
&\qquad = f\bigl(g(x)\bigr)\,g'(x).
\end{aligned} d x d F ( g ( x ) ) = F ′ ( g ( x ) ) g ′ ( x ) = f ( g ( x ) ) g ′ ( x ) . Read backward, an integrand of the form
f ( g ( x ) ) g ′ ( x ) f\bigl(g(x)\bigr)\,g'(x) f ( g ( x ) ) g ′ ( x ) has the antiderivative
F ( g ( x ) ) F\bigl(g(x)\bigr) F ( g ( x ) ) . Naming the inside function
u = g ( x ) u = g(x) u = g ( x ) turns it into an integral the table of basic integrals covers.
In practice, choose
u u u to be an inside function whose derivative is also a factor of the integrand. Write
d u du d u , rewrite the whole integral in terms of
u u u , integrate, and substitute
g ( x ) g(x) g ( x ) back for
u u u .
Example 1
Find each indefinite integral.
(a) ∫ 2 x ( x 2 + 5 ) 4 d x (b) ∫ e sin x cos x d x (c) ∫ 3 x 2 x 3 + 1 d x \begin{aligned}
&\text{(a)}\;\; \int 2x\,(x^2 + 5)^4\,dx \\[10pt]
&\text{(b)}\;\; \int e^{\sin x} \cos x\,dx \\[10pt]
&\text{(c)}\;\; \int \frac{3x^2}{\sqrt{x^3 + 1}}\,dx
\end{aligned} (a) ∫ 2 x ( x 2 + 5 ) 4 d x (b) ∫ e s i n x cos x d x (c) ∫ x 3 + 1 3 x 2 d x (a) Let
u = x 2 + 5 u = x^2 + 5 u = x 2 + 5 . Then
d u = 2 x d x du = 2x\,dx d u = 2 x d x , which is exactly the rest of the integrand:
∫ 2 x ( x 2 + 5 ) 4 d x = ∫ u 4 d u = u 5 5 + C = ( x 2 + 5 ) 5 5 + C . \begin{aligned}
&\int 2x\,(x^2 + 5)^4\,dx \\[4pt]
&\qquad = \int u^4\,du = \frac{u^5}{5} + C \\[4pt]
&\qquad = \frac{(x^2 + 5)^5}{5} + C.
\end{aligned} ∫ 2 x ( x 2 + 5 ) 4 d x = ∫ u 4 d u = 5 u 5 + C = 5 ( x 2 + 5 ) 5 + C . (b) Let
u = sin x u = \sin x u = sin x . Then
d u = cos x d x du = \cos x\,dx d u = cos x d x , and
∫ e sin x cos x d x = ∫ e u d u = e u + C = e sin x + C . \begin{aligned}
&\int e^{\sin x} \cos x\,dx \\[4pt]
&\qquad = \int e^u\,du = e^u + C \\[4pt]
&\qquad = e^{\sin x} + C.
\end{aligned} ∫ e s i n x cos x d x = ∫ e u d u = e u + C = e s i n x + C . (c) Let
u = x 3 + 1 u = x^3 + 1 u = x 3 + 1 . Then
d u = 3 x 2 d x du = 3x^2\,dx d u = 3 x 2 d x , and
∫ 3 x 2 x 3 + 1 d x = ∫ u − 1 / 2 d u = 2 u 1 / 2 + C = 2 x 3 + 1 + C . \begin{aligned}
&\int \frac{3x^2}{\sqrt{x^3 + 1}}\,dx \\[4pt]
&\qquad = \int u^{-1/2}\,du = 2u^{1/2} + C \\[4pt]
&\qquad = 2\sqrt{x^3 + 1} + C.
\end{aligned} ∫ x 3 + 1 3 x 2 d x = ∫ u − 1/2 d u = 2 u 1/2 + C = 2 x 3 + 1 + C . Adjusting a constant Often
d u du d u matches the rest of the integrand except for a constant factor. Since a constant factor can be taken outside an integral, multiply and divide by it. Only a constant can be adjusted this way. In
∫ cos ( x 2 ) d x \int \cos(x^2)\,dx ∫ cos ( x 2 ) d x , the choice
u = x 2 u = x^2 u = x 2 needs
d u = 2 x d x du = 2x\,dx d u = 2 x d x , and the integrand has no factor of
x x x to supply it.
Example 2
Find each indefinite integral.
(a) ∫ x cos ( x 2 ) d x (b) ∫ sin ( 5 x − 1 ) d x \begin{aligned}
&\text{(a)}\;\; \int x \cos(x^2)\,dx \\[10pt]
&\text{(b)}\;\; \int \sin(5x - 1)\,dx
\end{aligned} (a) ∫ x cos ( x 2 ) d x (b) ∫ sin ( 5 x − 1 ) d x (a) Let
u = x 2 u = x^2 u = x 2 , so
d u = 2 x d x du = 2x\,dx d u = 2 x d x and
x d x = 1 2 d u x\,dx = \frac{1}{2}\,du x d x = 2 1 d u :
∫ x cos ( x 2 ) d x = 1 2 ∫ cos u d u = 1 2 sin ( x 2 ) + C . \begin{aligned}
&\int x \cos(x^2)\,dx \\[4pt]
&\qquad = \frac{1}{2}\int \cos u\,du \\[4pt]
&\qquad = \frac{1}{2}\sin(x^2) + C.
\end{aligned} ∫ x cos ( x 2 ) d x = 2 1 ∫ cos u d u = 2 1 sin ( x 2 ) + C . (b) Let
u = 5 x − 1 u = 5x - 1 u = 5 x − 1 , so
d u = 5 d x du = 5\,dx d u = 5 d x and
d x = 1 5 d u dx = \frac{1}{5}\,du d x = 5 1 d u :
∫ sin ( 5 x − 1 ) d x = 1 5 ∫ sin u d u = − 1 5 cos ( 5 x − 1 ) + C . \begin{aligned}
&\int \sin(5x - 1)\,dx \\[4pt]
&\qquad = \frac{1}{5}\int \sin u\,du \\[4pt]
&\qquad = -\frac{1}{5}\cos(5x - 1) + C.
\end{aligned} ∫ sin ( 5 x − 1 ) d x = 5 1 ∫ sin u d u = − 5 1 cos ( 5 x − 1 ) + C . A leftover x Sometimes, after
g ( x ) g(x) g ( x ) and
d x dx d x are replaced, a factor of
x x x is left over. Solve
u = g ( x ) u = g(x) u = g ( x ) for
x x x and replace that too.
Example 3
Find the indefinite integral below.
∫ x x + 4 d x \int \frac{x}{\sqrt{x + 4}}\,dx ∫ x + 4 x d x Let
u = x + 4 u = x + 4 u = x + 4 , so
d u = d x du = dx d u = d x and
x = u − 4 x = u - 4 x = u − 4 . Then
∫ x x + 4 d x = ∫ u − 4 u 1 / 2 d u = ∫ ( u 1 / 2 − 4 u − 1 / 2 ) d u = 2 3 u 3 / 2 − 8 u 1 / 2 + C = 2 3 ( x + 4 ) 3 / 2 − 8 x + 4 + C . \begin{aligned}
&\int \frac{x}{\sqrt{x + 4}}\,dx \\[4pt]
&\qquad = \int \frac{u - 4}{u^{1/2}}\,du \\[4pt]
&\qquad = \int \left(u^{1/2} - 4u^{-1/2}\right)du \\[4pt]
&\qquad = \frac{2}{3}u^{3/2} - 8u^{1/2} + C \\[4pt]
&\qquad = \frac{2}{3}(x + 4)^{3/2} - 8\sqrt{x + 4} + C.
\end{aligned} ∫ x + 4 x d x = ∫ u 1/2 u − 4 d u = ∫ ( u 1/2 − 4 u − 1/2 ) d u = 3 2 u 3/2 − 8 u 1/2 + C = 3 2 ( x + 4 ) 3/2 − 8 x + 4 + C . Definite integrals The limits change with the variable: when
x x x runs from
a a a to
b b b ,
u u u starts at
g ( a ) g(a) g ( a ) and ends at
g ( b ) g(b) g ( b ) . If
F F F is an antiderivative of
f f f , both sides equal
F ( g ( b ) ) − F ( g ( a ) ) F\bigl(g(b)\bigr) - F\bigl(g(a)\bigr) F ( g ( b ) ) − F ( g ( a ) ) , so there's no need to substitute back.
Example 4
(a) ∫ 0 2 x ( x 2 + 1 ) 3 d x (b) ∫ 0 π / 2 cos 2 x sin x d x \begin{aligned}
&\text{(a)}\;\; \int_{0}^{2} x\,(x^2 + 1)^3\,dx \\[10pt]
&\text{(b)}\;\; \int_{0}^{\pi/2} \cos^2 x \sin x\,dx
\end{aligned} (a) ∫ 0 2 x ( x 2 + 1 ) 3 d x (b) ∫ 0 π /2 cos 2 x sin x d x (a) Let
u = x 2 + 1 u = x^2 + 1 u = x 2 + 1 , so
x d x = 1 2 d u x\,dx = \frac{1}{2}\,du x d x = 2 1 d u . When
x = 0 x = 0 x = 0 ,
u = 1 u = 1 u = 1 , and when
x = 2 x = 2 x = 2 ,
u = 5 u = 5 u = 5 . So
∫ 0 2 x ( x 2 + 1 ) 3 d x = 1 2 ∫ 1 5 u 3 d u = 1 8 u 4 ∣ 1 5 = 625 − 1 8 = 78. \begin{aligned}
&\int_{0}^{2} x\,(x^2 + 1)^3\,dx \\[4pt]
&\qquad = \frac{1}{2}\int_{1}^{5} u^3\,du = \frac{1}{8}\,u^4\,\Big|_{1}^{5} \\[4pt]
&\qquad = \frac{625 - 1}{8} = 78.
\end{aligned} ∫ 0 2 x ( x 2 + 1 ) 3 d x = 2 1 ∫ 1 5 u 3 d u = 8 1 u 4 1 5 = 8 625 − 1 = 78. (b) Let
u = cos x u = \cos x u = cos x , so
d u = − sin x d x du = -\sin x\,dx d u = − sin x d x . When
x = 0 x = 0 x = 0 ,
u = 1 u = 1 u = 1 , and when
x = π 2 x = \frac{\pi}{2} x = 2 π ,
u = 0 u = 0 u = 0 . The new limits run downward, and reversing them absorbs the minus sign:
∫ 0 π / 2 cos 2 x sin x d x = − ∫ 1 0 u 2 d u = ∫ 0 1 u 2 d u = 1 3 . \begin{aligned}
&\int_{0}^{\pi/2} \cos^2 x \sin x\,dx \\[4pt]
&\qquad = -\int_{1}^{0} u^2\,du \\[4pt]
&\qquad = \int_{0}^{1} u^2\,du = \frac{1}{3}.
\end{aligned} ∫ 0 π /2 cos 2 x sin x d x = − ∫ 1 0 u 2 d u = ∫ 0 1 u 2 d u = 3 1 . Logarithms from substitution When the integrand is a fraction whose numerator is the derivative of its denominator, the substitution
u = g ( x ) u = g(x) u = g ( x ) leaves
∫ 1 u d u \int \frac{1}{u}\,du ∫ u 1 d u :
The logarithm form
If
g g g has a continuous derivative and
g ( x ) ≠ 0 g(x) \ne 0 g ( x ) = 0 on an interval, then on that interval
∫ g ′ ( x ) g ( x ) d x = ln ∣ g ( x ) ∣ + C . \int \frac{g'(x)}{g(x)}\,dx = \ln\bigl|g(x)\bigr| + C. ∫ g ( x ) g ′ ( x ) d x = ln g ( x ) + C . Example 5
Find each indefinite integral.
(a) ∫ 2 x + 3 x 2 + 3 x + 7 d x (b) ∫ tan x d x (c) ∫ x x 2 + 9 d x \begin{aligned}
&\text{(a)}\;\; \int \frac{2x + 3}{x^2 + 3x + 7}\,dx \\[10pt]
&\text{(b)}\;\; \int \tan x\,dx \\[10pt]
&\text{(c)}\;\; \int \frac{x}{x^2 + 9}\,dx
\end{aligned} (a) ∫ x 2 + 3 x + 7 2 x + 3 d x (b) ∫ tan x d x (c) ∫ x 2 + 9 x d x (a) The numerator is the derivative of the denominator. Since
x 2 + 3 x + 7 = ( x + 3 2 ) 2 + 19 4 x^2 + 3x + 7 = \left(x + \frac{3}{2}\right)^2 + \frac{19}{4} x 2 + 3 x + 7 = ( x + 2 3 ) 2 + 4 19 , the denominator is always positive, so no absolute value is needed:
∫ 2 x + 3 x 2 + 3 x + 7 d x = ln ( x 2 + 3 x + 7 ) + C . \begin{aligned}
&\int \frac{2x + 3}{x^2 + 3x + 7}\,dx \\[4pt]
&\qquad = \ln\left(x^2 + 3x + 7\right) + C.
\end{aligned} ∫ x 2 + 3 x + 7 2 x + 3 d x = ln ( x 2 + 3 x + 7 ) + C . (b) Write
tan x = sin x cos x \tan x = \frac{\sin x}{\cos x} tan x = c o s x s i n x . The numerator is the derivative of
cos x \cos x cos x with the sign changed, so on any interval where
cos x ≠ 0 \cos x \ne 0 cos x = 0 ,
∫ tan x d x = − ∫ − sin x cos x d x = − ln ∣ cos x ∣ + C . \begin{aligned}
&\int \tan x\,dx \\[4pt]
&\qquad = -\int \frac{-\sin x}{\cos x}\,dx \\[4pt]
&\qquad = -\ln|\cos x| + C.
\end{aligned} ∫ tan x d x = − ∫ cos x − sin x d x = − ln ∣ cos x ∣ + C . (c) The derivative of
x 2 + 9 x^2 + 9 x 2 + 9 is
2 x 2x 2 x , so adjust the constant. Since
x 2 + 9 > 0 x^2 + 9 > 0 x 2 + 9 > 0 , no absolute value is needed:
∫ x x 2 + 9 d x = 1 2 ∫ 2 x x 2 + 9 d x = 1 2 ln ( x 2 + 9 ) + C . \begin{aligned}
&\int \frac{x}{x^2 + 9}\,dx \\[4pt]
&\qquad = \frac{1}{2}\int \frac{2x}{x^2 + 9}\,dx \\[4pt]
&\qquad = \frac{1}{2}\ln\left(x^2 + 9\right) + C.
\end{aligned} ∫ x 2 + 9 x d x = 2 1 ∫ x 2 + 9 2 x d x = 2 1 ln ( x 2 + 9 ) + C . From a table Example 6
The function
h h h has a continuous derivative, and the table below gives selected values of
h h h and
h ′ h' h ′ . Use it to evaluate each integral.
x x x 0 0 0 1 1 1 2 2 2 3 3 3 h ( x ) h(x) h ( x ) 5 5 5 − 1 -1 − 1 4 4 4 2 2 2 h ′ ( x ) h'(x) h ′ ( x ) − 4 -4 − 4 0 0 0 1 1 1 − 3 -3 − 3
(a) ∫ 1 3 h ′ ( x ) [ h ( x ) ] 2 d x (b) ∫ 0 1 h ′ ( 2 x ) d x \begin{aligned}
&\text{(a)}\;\; \int_{1}^{3} h'(x)\,\bigl[h(x)\bigr]^2\,dx \\[10pt]
&\text{(b)}\;\; \int_{0}^{1} h'(2x)\,dx
\end{aligned} (a) ∫ 1 3 h ′ ( x ) [ h ( x ) ] 2 d x (b) ∫ 0 1 h ′ ( 2 x ) d x (a) Let
u = h ( x ) u = h(x) u = h ( x ) , so
d u = h ′ ( x ) d x du = h'(x)\,dx d u = h ′ ( x ) d x . The limits become
h ( 1 ) = − 1 h(1) = -1 h ( 1 ) = − 1 and
h ( 3 ) = 2 h(3) = 2 h ( 3 ) = 2 :
∫ 1 3 h ′ ( x ) [ h ( x ) ] 2 d x = ∫ − 1 2 u 2 d u = 8 − ( − 1 ) 3 = 3. \begin{aligned}
&\int_{1}^{3} h'(x)\,\bigl[h(x)\bigr]^2\,dx \\[4pt]
&\qquad = \int_{-1}^{2} u^2\,du \\[4pt]
&\qquad = \frac{8 - (-1)}{3} = 3.
\end{aligned} ∫ 1 3 h ′ ( x ) [ h ( x ) ] 2 d x = ∫ − 1 2 u 2 d u = 3 8 − ( − 1 ) = 3. (b) Let
u = 2 x u = 2x u = 2 x , so
d x = 1 2 d u dx = \frac{1}{2}\,du d x = 2 1 d u , and the limits become
0 0 0 and
2 2 2 . An antiderivative of
h ′ h' h ′ is
h h h , so the values of
h ′ h' h ′ in the table aren't needed:
∫ 0 1 h ′ ( 2 x ) d x = 1 2 ∫ 0 2 h ′ ( u ) d u = 1 2 ( h ( 2 ) − h ( 0 ) ) = 1 2 ( 4 − 5 ) = − 1 2 . \begin{aligned}
&\int_{0}^{1} h'(2x)\,dx \\[4pt]
&\qquad = \frac{1}{2}\int_{0}^{2} h'(u)\,du \\[4pt]
&\qquad = \frac{1}{2}\bigl(h(2) - h(0)\bigr) \\[4pt]
&\qquad = \frac{1}{2}(4 - 5) = -\frac{1}{2}.
\end{aligned} ∫ 0 1 h ′ ( 2 x ) d x = 2 1 ∫ 0 2 h ′ ( u ) d u = 2 1 ( h ( 2 ) − h ( 0 ) ) = 2 1 ( 4 − 5 ) = − 2 1 . Long division and completing the square Some fractions need rewriting before a substitution fits. When the degree of the numerator is at least the degree of the denominator, divide first. The quotient is a polynomial, and the remainder over the divisor is a proper fraction, often one that a substitution finishes.
Example 7
Find each indefinite integral.
(a) ∫ x 2 + 3 x x − 2 d x (b) ∫ x 4 + 3 x 2 x 2 + 1 d x \begin{aligned}
&\text{(a)}\;\; \int \frac{x^2 + 3x}{x - 2}\,dx \\[10pt]
&\text{(b)}\;\; \int \frac{x^4 + 3x^2}{x^2 + 1}\,dx
\end{aligned} (a) ∫ x − 2 x 2 + 3 x d x (b) ∫ x 2 + 1 x 4 + 3 x 2 d x (a) Dividing gives
x 2 + 3 x = ( x − 2 ) ( x + 5 ) + 10 x^2 + 3x = (x - 2)(x + 5) + 10 x 2 + 3 x = ( x − 2 ) ( x + 5 ) + 10 , so
∫ x 2 + 3 x x − 2 d x = ∫ ( x + 5 + 10 x − 2 ) d x = x 2 2 + 5 x + 10 ln ∣ x − 2 ∣ + C . \begin{aligned}
&\int \frac{x^2 + 3x}{x - 2}\,dx \\[4pt]
&\qquad = \int \left(x + 5 + \frac{10}{x - 2}\right)dx \\[4pt]
&\qquad = \frac{x^2}{2} + 5x + 10\ln|x - 2| + C.
\end{aligned} ∫ x − 2 x 2 + 3 x d x = ∫ ( x + 5 + x − 2 10 ) d x = 2 x 2 + 5 x + 10 ln ∣ x − 2∣ + C . (b) Dividing gives
x 4 + 3 x 2 = ( x 2 + 1 ) ( x 2 + 2 ) − 2 x^4 + 3x^2 = \left(x^2 + 1\right)\left(x^2 + 2\right) - 2 x 4 + 3 x 2 = ( x 2 + 1 ) ( x 2 + 2 ) − 2 . The remainder over the divisor is a multiple of the arctangent form:
∫ x 4 + 3 x 2 x 2 + 1 d x = ∫ ( x 2 + 2 − 2 x 2 + 1 ) d x = x 3 3 + 2 x − 2 arctan x + C . \begin{aligned}
&\int \frac{x^4 + 3x^2}{x^2 + 1}\,dx \\[4pt]
&\qquad = \int \left(x^2 + 2 - \frac{2}{x^2 + 1}\right)dx \\[4pt]
&\qquad = \frac{x^3}{3} + 2x - 2\arctan x + C.
\end{aligned} ∫ x 2 + 1 x 4 + 3 x 2 d x = ∫ ( x 2 + 2 − x 2 + 1 2 ) d x = 3 x 3 + 2 x − 2 arctan x + C . When a quadratic in a denominator doesn't factor, or sits under a square root, complete the square. A substitution then turns it into one of two inverse trigonometric forms.
Example 8
(a) ∫ d x x 2 − 2 x + 17 (b) ∫ 2 9 / 2 d x 21 + 4 x − x 2 (c) ∫ x + 7 x 2 + 4 x + 13 d x \begin{aligned}
&\text{(a)}\;\; \int \frac{dx}{x^2 - 2x + 17} \\[10pt]
&\text{(b)}\;\; \int_{2}^{9/2} \frac{dx}{\sqrt{21 + 4x - x^2}} \\[10pt]
&\text{(c)}\;\; \int \frac{x + 7}{x^2 + 4x + 13}\,dx
\end{aligned} (a) ∫ x 2 − 2 x + 17 d x (b) ∫ 2 9/2 21 + 4 x − x 2 d x (c) ∫ x 2 + 4 x + 13 x + 7 d x (a) Complete the square:
x 2 − 2 x + 17 = ( x − 1 ) 2 + 16 x^2 - 2x + 17 = (x - 1)^2 + 16 x 2 − 2 x + 17 = ( x − 1 ) 2 + 16 . With
u = x − 1 u = x - 1 u = x − 1 , so
d u = d x du = dx d u = d x , and
a = 4 a = 4 a = 4 ,
∫ d x ( x − 1 ) 2 + 16 = 1 4 arctan x − 1 4 + C . \begin{aligned}
&\int \frac{dx}{(x - 1)^2 + 16} \\[4pt]
&\qquad = \frac{1}{4}\arctan\frac{x - 1}{4} + C.
\end{aligned} ∫ ( x − 1 ) 2 + 16 d x = 4 1 arctan 4 x − 1 + C . (b) Take out the minus sign before completing the square.
21 + 4 x − x 2 = 21 − ( x 2 − 4 x ) = 21 + 4 − ( x 2 − 4 x + 4 ) = 25 − ( x − 2 ) 2 \begin{aligned}
&21 + 4x - x^2 \\[4pt]
&\qquad = 21 - \left(x^2 - 4x\right) \\[4pt]
&\qquad = 21 + 4 - \left(x^2 - 4x + 4\right) \\[4pt]
&\qquad = 25 - (x - 2)^2
\end{aligned} 21 + 4 x − x 2 = 21 − ( x 2 − 4 x ) = 21 + 4 − ( x 2 − 4 x + 4 ) = 25 − ( x − 2 ) 2 With
u = x − 2 u = x - 2 u = x − 2 , so
d u = d x du = dx d u = d x , and
a = 5 a = 5 a = 5 , the limits become
u = 0 u = 0 u = 0 and
u = 5 2 u = \frac{5}{2} u = 2 5 :
∫ 2 9 / 2 d x 25 − ( x − 2 ) 2 = ∫ 0 5 / 2 d u 25 − u 2 = arcsin u 5 ∣ 0 5 / 2 = arcsin 1 2 − arcsin 0 = π 6 . \begin{aligned}
&\int_{2}^{9/2} \frac{dx}{\sqrt{25 - (x - 2)^2}} \\[4pt]
&\qquad = \int_{0}^{5/2} \frac{du}{\sqrt{25 - u^2}} \\[4pt]
&\qquad = \arcsin\frac{u}{5}\,\Big|_{0}^{5/2} \\[4pt]
&\qquad = \arcsin\frac{1}{2} - \arcsin 0 = \frac{\pi}{6}.
\end{aligned} ∫ 2 9/2 25 − ( x − 2 ) 2 d x = ∫ 0 5/2 25 − u 2 d u = arcsin 5 u 0 5/2 = arcsin 2 1 − arcsin 0 = 6 π . (c) The derivative of the denominator is
2 x + 4 2x + 4 2 x + 4 . Write the numerator as half of that plus what's left,
x + 7 = 1 2 ( 2 x + 4 ) + 5 x + 7 = \frac{1}{2}(2x + 4) + 5 x + 7 = 2 1 ( 2 x + 4 ) + 5 , and split the integral. The first piece is the logarithm form. For the second,
x 2 + 4 x + 13 = ( x + 2 ) 2 + 9 x^2 + 4x + 13 = (x + 2)^2 + 9 x 2 + 4 x + 13 = ( x + 2 ) 2 + 9 , with
u = x + 2 u = x + 2 u = x + 2 ,
d u = d x du = dx d u = d x , and
a = 3 a = 3 a = 3 :
∫ x + 7 x 2 + 4 x + 13 d x = 1 2 ∫ 2 x + 4 x 2 + 4 x + 13 d x + ∫ 5 d x ( x + 2 ) 2 + 9 = 1 2 ln ( x 2 + 4 x + 13 ) + 5 3 arctan x + 2 3 + C . \begin{aligned}
&\int \frac{x + 7}{x^2 + 4x + 13}\,dx \\[4pt]
&\qquad = \frac{1}{2}\int \frac{2x + 4}{x^2 + 4x + 13}\,dx \\[4pt]
&\qquad\qquad + \int \frac{5\,dx}{(x + 2)^2 + 9} \\[4pt]
&\qquad = \frac{1}{2}\ln\left(x^2 + 4x + 13\right) \\[4pt]
&\qquad\qquad + \frac{5}{3}\arctan\frac{x + 2}{3} + C.
\end{aligned} ∫ x 2 + 4 x + 13 x + 7 d x = 2 1 ∫ x 2 + 4 x + 13 2 x + 4 d x + ∫ ( x + 2 ) 2 + 9 5 d x = 2 1 ln ( x 2 + 4 x + 13 ) + 3 5 arctan 3 x + 2 + C .