Each technique in this unit fits a particular form of integrand. Most of the work in a new integral is recognizing which form it has, and the questions below are a reasonable order in which to ask.
A checklist What you see What to try A basic form, or one after expanding, splitting or simplifying The basic rules A factor that is the derivative of an inside function, up to a constant Substitution A fraction whose numerator has degree at least the denominator's Long division, then look again A quadratic denominator with no real zeros Completing the square, for an arctangent, after splitting off a logarithm if the numerator has an x x x A denominator with distinct linear factors Partial fractions A product of unlike functions, or a logarithm or inverse trigonometric function alone Integration by parts: repeated, or by a table, when u u u
Check the substitution row early. A fraction whose numerator is a multiple of the denominator's derivative needs no partial fractions and no completing the square.
Rewrites that are legal Example 1
Decide which of these are equal to
∫ 1 4 ( x + 1 ) 2 x d x \int_1^4 \frac{(x + 1)^2}{\sqrt{x}}\,dx ∫ 1 4 x ( x + 1 ) 2 d x , and use one of them to evaluate it.
I. ∫ 1 4 ( x 3 / 2 + 2 x 1 / 2 + x − 1 / 2 ) d x II. ∫ 1 4 ( x + 1 ) 2 d x ∫ 1 4 x d x III. ∫ 1 2 2 ( u 2 + 1 ) 2 d u \begin{aligned}
&\text{I.}\;\; \int_1^4 \left(x^{3/2} + 2x^{1/2} + x^{-1/2}\right)dx \\[10pt]
&\text{II.}\;\; \frac{\int_1^4 (x + 1)^2\,dx}{\int_1^4 \sqrt{x}\,dx} \\[10pt]
&\text{III.}\;\; \int_1^2 2\left(u^2 + 1\right)^2\,du
\end{aligned} I. ∫ 1 4 ( x 3/2 + 2 x 1/2 + x − 1/2 ) d x II. ∫ 1 4 x d x ∫ 1 4 ( x + 1 ) 2 d x III. ∫ 1 2 2 ( u 2 + 1 ) 2 d u I is equal: expand
( x + 1 ) 2 = x 2 + 2 x + 1 (x + 1)^2 = x^2 + 2x + 1 ( x + 1 ) 2 = x 2 + 2 x + 1 and divide each term by
x 1 / 2 x^{1/2} x 1/2 .
II isn't. The integral of a quotient is not the quotient of the integrals. It works out to
39 14 / 3 = 117 14 \frac{39}{14/3} = \frac{117}{14} 14/3 39 = 14 117 , which differs from the true value.
III is equal. With
u = x u = \sqrt{x} u = x ,
x = u 2 x = u^2 x = u 2 , so
d x = 2 u d u dx = 2u\,du d x = 2 u d u and
( x + 1 ) 2 = ( u 2 + 1 ) 2 (x + 1)^2 = (u^2 + 1)^2 ( x + 1 ) 2 = ( u 2 + 1 ) 2 . The factor
1 x = 1 u \frac{1}{\sqrt{x}} = \frac{1}{u} x 1 = u 1 cancels the
u u u in
d x dx d x , and the limits become
u = 1 u = 1 u = 1 and
u = 2 u = 2 u = 2 .
∫ 1 4 ( x 3 / 2 + 2 x 1 / 2 + x − 1 / 2 ) d x = [ 2 5 x 5 / 2 + 4 3 x 3 / 2 + 2 x 1 / 2 ] 1 4 = ( 64 5 + 32 3 + 4 ) − ( 2 5 + 4 3 + 2 ) = 356 15 \begin{aligned}
&\int_1^4 \left(x^{3/2} + 2x^{1/2} + x^{-1/2}\right)dx \\[4pt]
&\quad = \left[\frac{2}{5}x^{5/2} + \frac{4}{3}x^{3/2} + 2x^{1/2}\right]_1^4 \\[4pt]
&\quad = \left(\frac{64}{5} + \frac{32}{3} + 4\right) \\[4pt]
&\qquad - \left(\frac{2}{5} + \frac{4}{3} + 2\right) \\[4pt]
&\quad = \frac{356}{15}
\end{aligned} ∫ 1 4 ( x 3/2 + 2 x 1/2 + x − 1/2 ) d x = [ 5 2 x 5/2 + 3 4 x 3/2 + 2 x 1/2 ] 1 4 = ( 5 64 + 3 32 + 4 ) − ( 5 2 + 3 4 + 2 ) = 15 356 Three quadratic denominators Example 2
Name a technique for each integral, and find it.
(a) ∫ 3 x 2 − 8 x + 12 d x (b) ∫ 3 x 2 − 8 x + 16 d x (c) ∫ 3 x 2 − 8 x + 25 d x \begin{aligned}
&\text{(a)}\;\; \int \frac{3}{x^2 - 8x + 12}\,dx \\[10pt]
&\text{(b)}\;\; \int \frac{3}{x^2 - 8x + 16}\,dx \\[10pt]
&\text{(c)}\;\; \int \frac{3}{x^2 - 8x + 25}\,dx
\end{aligned} (a) ∫ x 2 − 8 x + 12 3 d x (b) ∫ x 2 − 8 x + 16 3 d x (c) ∫ x 2 − 8 x + 25 3 d x The discriminant
b 2 − 4 a c b^2 - 4ac b 2 − 4 a c of each denominator decides: positive gives two linear factors, zero gives a perfect square, and negative gives no real zeros.
(a) Partial fractions. The discriminant is
64 − 48 > 0 64 - 48 > 0 64 − 48 > 0 , and
x 2 − 8 x + 12 = ( x − 2 ) ( x − 6 ) x^2 - 8x + 12 = (x - 2)(x - 6) x 2 − 8 x + 12 = ( x − 2 ) ( x − 6 ) . Write
3 ( x − 2 ) ( x − 6 ) = A x − 2 + B x − 6 \frac{3}{(x - 2)(x - 6)} = \frac{A}{x - 2} + \frac{B}{x - 6} ( x − 2 ) ( x − 6 ) 3 = x − 2 A + x − 6 B . The basic equation
3 = A ( x − 6 ) + B ( x − 2 ) 3 = A(x - 6) + B(x - 2) 3 = A ( x − 6 ) + B ( x − 2 ) gives
A = − 3 4 A = -\frac{3}{4} A = − 4 3 at
x = 2 x = 2 x = 2 and
B = 3 4 B = \frac{3}{4} B = 4 3 at
x = 6 x = 6 x = 6 :
∫ 3 x 2 − 8 x + 12 d x = 3 4 ln ∣ x − 6 ∣ − 3 4 ln ∣ x − 2 ∣ + C . \begin{aligned}
&\int \frac{3}{x^2 - 8x + 12}\,dx \\[4pt]
&\quad = \frac{3}{4}\ln|x - 6| - \frac{3}{4}\ln|x - 2| + C.
\end{aligned} ∫ x 2 − 8 x + 12 3 d x = 4 3 ln ∣ x − 6∣ − 4 3 ln ∣ x − 2∣ + C . (b) Substitution. The discriminant is 0, and the denominator is
( x − 4 ) 2 (x - 4)^2 ( x − 4 ) 2 . With
u = x − 4 u = x - 4 u = x − 4 ,
d u = d x du = dx d u = d x :
∫ 3 ( x − 4 ) 2 d x = − 3 x − 4 + C . \int \frac{3}{(x - 4)^2}\,dx = -\frac{3}{x - 4} + C. ∫ ( x − 4 ) 2 3 d x = − x − 4 3 + C . (c) Completing the square. The discriminant is
64 − 100 < 0 64 - 100 < 0 64 − 100 < 0 , and
x 2 − 8 x + 25 = ( x − 4 ) 2 + 9 x^2 - 8x + 25 = (x - 4)^2 + 9 x 2 − 8 x + 25 = ( x − 4 ) 2 + 9 . With
u = x − 4 u = x - 4 u = x − 4 ,
d u = d x du = dx d u = d x , and
a = 3 a = 3 a = 3 :
∫ 3 ( x − 4 ) 2 + 9 d x = arctan x − 4 3 + C . \begin{aligned}
&\int \frac{3}{(x - 4)^2 + 9}\,dx \\[4pt]
&\quad = \arctan\frac{x - 4}{3} + C.
\end{aligned} ∫ ( x − 4 ) 2 + 9 3 d x = arctan 3 x − 4 + C . Products Example 3
Name a technique for each integral, and find it.
(a) ∫ ( 2 x + 1 ) 4 x 2 + x d x (b) ∫ x ⋅ 3 x d x (c) ∫ x arctan ( x 2 ) d x \begin{aligned}
&\text{(a)}\;\; \int (2x + 1)\,4^{x^2 + x}\,dx \\[10pt]
&\text{(b)}\;\; \int x \cdot 3^x\,dx \\[10pt]
&\text{(c)}\;\; \int x\arctan\left(x^2\right)dx
\end{aligned} (a) ∫ ( 2 x + 1 ) 4 x 2 + x d x (b) ∫ x ⋅ 3 x d x (c) ∫ x arctan ( x 2 ) d x (a) Substitution. The derivative of the exponent
x 2 + x x^2 + x x 2 + x is
2 x + 1 2x + 1 2 x + 1 , the other factor. With
u = x 2 + x u = x^2 + x u = x 2 + x ,
d u = ( 2 x + 1 ) d x du = (2x + 1)\,dx d u = ( 2 x + 1 ) d x , and
∫ 4 u d u = 4 u ln 4 \int 4^u\,du = \frac{4^u}{\ln 4} ∫ 4 u d u = l n 4 4 u :
∫ ( 2 x + 1 ) 4 x 2 + x d x = 4 x 2 + x ln 4 + C . \int (2x + 1)\,4^{x^2 + x}\,dx = \frac{4^{x^2 + x}}{\ln 4} + C. ∫ ( 2 x + 1 ) 4 x 2 + x d x = ln 4 4 x 2 + x + C . (b) Parts. Here
x x x isn't a multiple of the exponent's derivative, so no substitution fits. Take
u = x u = x u = x and
d v = 3 x d x dv = 3^x\,dx d v = 3 x d x , so
v = 3 x ln 3 v = \frac{3^x}{\ln 3} v = l n 3 3 x :
∫ x ⋅ 3 x d x = x ⋅ 3 x ln 3 − ∫ 3 x ln 3 d x = x ⋅ 3 x ln 3 − 3 x ( ln 3 ) 2 + C . \begin{aligned}
\int x \cdot 3^x\,dx &= \frac{x \cdot 3^x}{\ln 3} - \int \frac{3^x}{\ln 3}\,dx \\[4pt]
&= \frac{x \cdot 3^x}{\ln 3} - \frac{3^x}{(\ln 3)^2} + C.
\end{aligned} ∫ x ⋅ 3 x d x = ln 3 x ⋅ 3 x − ∫ ln 3 3 x d x = ln 3 x ⋅ 3 x − ( ln 3 ) 2 3 x + C . (c) Parts, with the inverse trigonometric factor as
u u u :
u = arctan ( x 2 ) u = \arctan\left(x^2\right) u = arctan ( x 2 ) ,
d v = x d x dv = x\,dx d v = x d x , so
d u = 2 x d x 1 + x 4 du = \frac{2x\,dx}{1 + x^4} d u = 1 + x 4 2 x d x and
v = x 2 2 v = \frac{x^2}{2} v = 2 x 2 . The leftover integral is a substitution, with
w = 1 + x 4 w = 1 + x^4 w = 1 + x 4 :
∫ x arctan ( x 2 ) d x = x 2 2 arctan ( x 2 ) − ∫ x 3 1 + x 4 d x = x 2 2 arctan ( x 2 ) − 1 4 ln ( 1 + x 4 ) + C . \begin{aligned}
&\int x\arctan\left(x^2\right)dx \\[4pt]
&\quad = \frac{x^2}{2}\arctan\left(x^2\right) - \int \frac{x^3}{1 + x^4}\,dx \\[4pt]
&\quad = \frac{x^2}{2}\arctan\left(x^2\right) \\[4pt]
&\qquad - \frac{1}{4}\ln\left(1 + x^4\right) + C.
\end{aligned} ∫ x arctan ( x 2 ) d x = 2 x 2 arctan ( x 2 ) − ∫ 1 + x 4 x 3 d x = 2 x 2 arctan ( x 2 ) − 4 1 ln ( 1 + x 4 ) + C .