When the basic table is not enough#
Sooner or later you meet an integral that none of the basic formulas fit. Usually it matches a special form, or it gives way to a special method. We will learn the standard forms involving x 2 ± a 2 x^2 \pm a^2 x 2 ± a 2 ∫ e x ( f + f ′ ) d x \displaystyle \int e^x(f + f')\,dx ∫ e x ( f + f ′ ) d x
∫ d x x 2 − a 2 = 1 2 a log ∣ x − a x + a ∣ , ∫ d x a 2 − x 2 = 1 2 a log ∣ a + x a − x ∣ , ∫ d x x 2 + a 2 = 1 a tan − 1 x a \displaystyle \int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left\lvert\frac{x - a}{x + a}\right\rvert, \quad \int\frac{dx}{a^2 - x^2} = \frac{1}{2a}\log\left\lvert\frac{a + x}{a - x}\right\rvert, \quad \int\frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} ∫ x 2 − a 2 d x = 2 a 1 log x + a x − a , ∫ a 2 − x 2 d x = 2 a 1 log a − x a + x , ∫ x 2 + a 2 d x = a 1 tan − 1 a x ∫ d x x 2 ± a 2 = log ∣ x + x 2 ± a 2 ∣ , ∫ d x a 2 − x 2 = sin − 1 x a \displaystyle \int\frac{dx}{\sqrt{x^2 \pm a^2}} = \log\left\lvert x + \sqrt{x^2 \pm a^2}\right\rvert, \quad \int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} ∫ x 2 ± a 2 d x = log x + x 2 ± a 2 , ∫ a 2 − x 2 d x = sin − 1 a x ∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 sin − 1 x a \displaystyle \int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} ∫ a 2 − x 2 d x = 2 x a 2 − x 2 + 2 a 2 sin − 1 a x
Add + C + C + C
Example 1. ∫ d x x 2 + 6 x + 13 = ∫ d x ( x + 3 ) 2 + 4 = 1 2 tan − 1 x + 3 2 + C \displaystyle \int\frac{dx}{x^2 + 6x + 13} = \int\frac{dx}{(x + 3)^2 + 4} = \tfrac{1}{2}\tan^{-1}\tfrac{x + 3}{2} + C ∫ x 2 + 6 x + 13 d x = ∫ ( x + 3 ) 2 + 4 d x = 2 1 tan − 1 2 x + 3 + C
Example 2. ∫ d x 8 + 2 x − x 2 = ∫ d x 9 − ( x − 1 ) 2 = sin − 1 x − 1 3 + C \displaystyle \int\frac{dx}{\sqrt{8 + 2x - x^2}} = \int\frac{dx}{\sqrt{9 - (x - 1)^2}} = \sin^{-1}\tfrac{x - 1}{3} + C ∫ 8 + 2 x − x 2 d x = ∫ 9 − ( x − 1 ) 2 d x = sin − 1 3 x − 1 + C
Example 3. ∫ d x 9 x 2 − 4 = 1 9 ∫ d x x 2 − ( 2 / 3 ) 2 = 1 12 log ∣ 3 x − 2 3 x + 2 ∣ + C \displaystyle \int\frac{dx}{9x^2 - 4} = \tfrac{1}{9}\int\frac{dx}{x^2 - (2/3)^2} = \tfrac{1}{12}\log\left\lvert\tfrac{3x - 2}{3x + 2}\right\rvert + C ∫ 9 x 2 − 4 d x = 9 1 ∫ x 2 − ( 2/3 ) 2 d x = 12 1 log 3 x + 2 3 x − 2 + C
Example 4. ∫ 2 x + 1 x 2 + 4 x + 8 d x \displaystyle \int\frac{2x + 1}{x^2 + 4x + 8}\,dx ∫ x 2 + 4 x + 8 2 x + 1 d x 2 x + 1 = ( 2 x + 4 ) − 3 2x + 1 = (2x + 4) - 3 2 x + 1 = ( 2 x + 4 ) − 3 log ( x 2 + 4 x + 8 ) − 3 2 tan − 1 x + 2 2 + C \displaystyle \log(x^2 + 4x + 8) - \tfrac{3}{2}\tan^{-1}\tfrac{x + 2}{2} + C log ( x 2 + 4 x + 8 ) − 2 3 tan − 1 2 x + 2 + C
Partial fractions#
For a proper rational function, break it up according to the factors of the denominator:
p x + q ( x − a ) ( x − b ) = A x − a + B x − b \displaystyle \frac{px + q}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b} ( x − a ) ( x − b ) p x + q = x − a A + x − b B p x + q ( x − a ) 2 = A x − a + B ( x − a ) 2 \displaystyle \frac{px + q}{(x - a)^2} = \frac{A}{x - a} + \frac{B}{(x - a)^2} ( x − a ) 2 p x + q = x − a A + ( x − a ) 2 B p x 2 + q x + r ( x − a ) ( x 2 + b x + c ) = A x − a + B x + C x 2 + b x + c \displaystyle \frac{px^2 + qx + r}{(x - a)(x^2 + bx + c)} = \frac{A}{x - a} + \frac{Bx + C}{x^2 + bx + c} ( x − a ) ( x 2 + b x + c ) p x 2 + q x + r = x − a A + x 2 + b x + c B x + C
One warning: if the degree of the numerator is not less than that of the denominator, do the long division first.
Example 5. ∫ 5 x − 1 ( x − 1 ) ( x + 2 ) d x \displaystyle \int\frac{5x - 1}{(x - 1)(x + 2)}\,dx ∫ ( x − 1 ) ( x + 2 ) 5 x − 1 d x 5 x − 1 = A ( x + 2 ) + B ( x − 1 ) 5x - 1 = A(x + 2) + B(x - 1) 5 x − 1 = A ( x + 2 ) + B ( x − 1 ) x = 1 x = 1 x = 1 A = 4 3 \displaystyle A = \tfrac{4}{3} A = 3 4 x = − 2 x = -2 x = − 2 B = 11 3 \displaystyle B = \tfrac{11}{3} B = 3 11 4 3 log ∣ x − 1 ∣ + 11 3 log ∣ x + 2 ∣ + C \displaystyle \tfrac{4}{3}\log\lvert x - 1 \rvert + \tfrac{11}{3}\log\lvert x + 2 \rvert + C 3 4 log ∣ x − 1 ∣ + 3 11 log ∣ x + 2 ∣ + C
Example 6. ∫ x ( x − 2 ) 2 d x = ∫ ( 1 x − 2 + 2 ( x − 2 ) 2 ) d x = log ∣ x − 2 ∣ − 2 x − 2 + C \displaystyle \int\frac{x}{(x - 2)^2}\,dx = \int\left(\frac{1}{x - 2} + \frac{2}{(x - 2)^2}\right)dx = \log\lvert x - 2 \rvert - \tfrac{2}{x - 2} + C ∫ ( x − 2 ) 2 x d x = ∫ ( x − 2 1 + ( x − 2 ) 2 2 ) d x = log ∣ x − 2 ∣ − x − 2 2 + C
Integration by parts#
∫ u v d x = u ∫ v d x − ∫ ( u ′ ∫ v d x ) d x . \displaystyle \int u\,v\,dx = u\int v\,dx - \int\left(u'\int v\,dx\right)dx. ∫ u v d x = u ∫ v d x − ∫ ( u ′ ∫ v d x ) d x .
Which function should be u u u ILATE , Inverse trig, Log, Algebraic, Trig, Exponential, and pick whichever comes first.
Example 7. ∫ x e 2 x d x = x e 2 x 2 − ∫ e 2 x 2 d x = e 2 x 4 ( 2 x − 1 ) + C \displaystyle \int x e^{2x}\,dx = \tfrac{xe^{2x}}{2} - \int\tfrac{e^{2x}}{2}dx = \tfrac{e^{2x}}{4}(2x - 1) + C ∫ x e 2 x d x = 2 x e 2 x − ∫ 2 e 2 x d x = 4 e 2 x ( 2 x − 1 ) + C
Example 8. ∫ x 2 log x d x = x 3 3 log x − ∫ x 2 3 d x = x 3 3 log x − x 3 9 + C \displaystyle \int x^2\log x\,dx = \tfrac{x^3}{3}\log x - \int\tfrac{x^2}{3}\,dx = \tfrac{x^3}{3}\log x - \tfrac{x^3}{9} + C ∫ x 2 log x d x = 3 x 3 log x − ∫ 3 x 2 d x = 3 x 3 log x − 9 x 3 + C
Example 9. ∫ tan − 1 x d x = x tan − 1 x − 1 2 log ( 1 + x 2 ) + C \displaystyle \int\tan^{-1}x\,dx = x\tan^{-1}x - \tfrac{1}{2}\log(1 + x^2) + C ∫ tan − 1 x d x = x tan − 1 x − 2 1 log ( 1 + x 2 ) + C
A shortcut worth remembering. ∫ e x ( f ( x ) + f ′ ( x ) ) d x = e x f ( x ) + C \displaystyle \int e^x(f(x) + f'(x))\,dx = e^x f(x) + C ∫ e x ( f ( x ) + f ′ ( x )) d x = e x f ( x ) + C ∫ e x ( 1 x − 1 x 2 ) d x = e x x + C \displaystyle \int e^x\left(\tfrac{1}{x} - \tfrac{1}{x^2}\right)dx = \tfrac{e^x}{x} + C ∫ e x ( x 1 − x 2 1 ) d x = x e x + C ∫ e x ( sin x + cos x ) d x = e x sin x + C \displaystyle \int e^x(\sin x + \cos x)\,dx = e^x\sin x + C ∫ e x ( sin x + cos x ) d x = e x sin x + C
Try these yourself#
∫ d x x 2 − 25 \displaystyle \int\frac{dx}{x^2 - 25} ∫ x 2 − 25 d x ∫ d x 4 + 9 x 2 \displaystyle \int\frac{dx}{4 + 9x^2} ∫ 4 + 9 x 2 d x ∫ d x x 2 + 16 \displaystyle \int\frac{dx}{\sqrt{x^2 + 16}} ∫ x 2 + 16 d x ∫ d x x 2 − 4 x + 13 \displaystyle \int\frac{dx}{x^2 - 4x + 13} ∫ x 2 − 4 x + 13 d x ∫ d x 5 − 4 x − x 2 \displaystyle \int\frac{dx}{\sqrt{5 - 4x - x^2}} ∫ 5 − 4 x − x 2 d x ∫ 3 x − 2 x 2 + 2 x + 5 d x \displaystyle \int\frac{3x - 2}{x^2 + 2x + 5}\,dx ∫ x 2 + 2 x + 5 3 x − 2 d x ∫ x + 4 x 2 − 3 x + 2 d x \displaystyle \int\frac{x + 4}{x^2 - 3x + 2}\,dx ∫ x 2 − 3 x + 2 x + 4 d x ∫ 2 x ( x + 1 ) ( x 2 + 1 ) d x \displaystyle \int\frac{2x}{(x + 1)(x^2 + 1)}\,dx ∫ ( x + 1 ) ( x 2 + 1 ) 2 x d x ∫ x sin 3 x d x \displaystyle \int x\sin 3x\,dx ∫ x sin 3 x d x ∫ x log ( 2 x ) d x \displaystyle \int x\log(2x)\,dx ∫ x log ( 2 x ) d x ∫ x 2 e − x d x \displaystyle \int x^2 e^{-x}\,dx ∫ x 2 e − x d x ∫ sin − 1 x d x \displaystyle \int\sin^{-1}x\,dx ∫ sin − 1 x d x ∫ e x ( tan x + sec 2 x ) d x \displaystyle \int e^x\left(\tan x + \sec^2 x\right)dx ∫ e x ( tan x + sec 2 x ) d x ∫ 16 − x 2 d x \displaystyle \int\sqrt{16 - x^2}\,dx ∫ 16 − x 2 d x ∫ e 2 x sin x d x \displaystyle \int e^{2x}\sin x\,dx ∫ e 2 x sin x d x
Answers to check against#
Show answers
1 10 log ∣ x − 5 x + 5 ∣ + C \displaystyle \tfrac{1}{10}\log\left\lvert\tfrac{x - 5}{x + 5}\right\rvert + C 10 1 log x + 5 x − 5 + C 1 6 tan − 1 3 x 2 + C \displaystyle \tfrac{1}{6}\tan^{-1}\tfrac{3x}{2} + C 6 1 tan − 1 2 3 x + C log ∣ x + x 2 + 16 ∣ + C \log\lvert x + \sqrt{x^2 + 16}\rvert + C log ∣ x + x 2 + 16 ∣ + C 1 3 tan − 1 x − 2 3 + C \displaystyle \tfrac{1}{3}\tan^{-1}\tfrac{x - 2}{3} + C 3 1 tan − 1 3 x − 2 + C sin − 1 x + 2 3 + C \displaystyle \sin^{-1}\tfrac{x + 2}{3} + C sin − 1 3 x + 2 + C 3 2 log ( x 2 + 2 x + 5 ) − 5 2 tan − 1 x + 1 2 + C \displaystyle \tfrac{3}{2}\log(x^2 + 2x + 5) - \tfrac{5}{2}\tan^{-1}\tfrac{x + 1}{2} + C 2 3 log ( x 2 + 2 x + 5 ) − 2 5 tan − 1 2 x + 1 + C x + 4 ( x − 1 ) ( x − 2 ) = − 5 x − 1 + 6 x − 2 \displaystyle \tfrac{x + 4}{(x - 1)(x - 2)} = \tfrac{-5}{x - 1} + \tfrac{6}{x - 2} ( x − 1 ) ( x − 2 ) x + 4 = x − 1 − 5 + x − 2 6 6 log ∣ x − 2 ∣ − 5 log ∣ x − 1 ∣ + C 6\log\lvert x - 2\rvert - 5\log\lvert x - 1\rvert + C 6 log ∣ x − 2 ∣ − 5 log ∣ x − 1 ∣ + C 2 x ( x + 1 ) ( x 2 + 1 ) = − 1 x + 1 + x + 1 x 2 + 1 \displaystyle \tfrac{2x}{(x + 1)(x^2 + 1)} = \tfrac{-1}{x + 1} + \tfrac{x + 1}{x^2 + 1} ( x + 1 ) ( x 2 + 1 ) 2 x = x + 1 − 1 + x 2 + 1 x + 1 − log ∣ x + 1 ∣ + 1 2 log ( x 2 + 1 ) + tan − 1 x + C \displaystyle -\log\lvert x + 1\rvert + \tfrac{1}{2}\log(x^2 + 1) + \tan^{-1}x + C − log ∣ x + 1 ∣ + 2 1 log ( x 2 + 1 ) + tan − 1 x + C − x cos 3 x 3 + sin 3 x 9 + C \displaystyle -\tfrac{x\cos 3x}{3} + \tfrac{\sin 3x}{9} + C − 3 x c o s 3 x + 9 s i n 3 x + C x 2 2 log 2 x − x 2 4 + C \displaystyle \tfrac{x^2}{2}\log 2x - \tfrac{x^2}{4} + C 2 x 2 log 2 x − 4 x 2 + C − e − x ( x 2 + 2 x + 2 ) + C -e^{-x}(x^2 + 2x + 2) + C − e − x ( x 2 + 2 x + 2 ) + C x sin − 1 x + 1 − x 2 + C x\sin^{-1}x + \sqrt{1 - x^2} + C x sin − 1 x + 1 − x 2 + C e x tan x + C e^x\tan x + C e x tan x + C x 2 16 − x 2 + 8 sin − 1 x 4 + C \displaystyle \tfrac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\tfrac{x}{4} + C 2 x 16 − x 2 + 8 sin − 1 4 x + C e 2 x 5 ( 2 sin x − cos x ) + C \displaystyle \tfrac{e^{2x}}{5}(2\sin x - \cos x) + C 5 e 2 x ( 2 sin x − cos x ) + C