Integration by Parts
Integration by parts is a technique based on the product rule for differentiation. The formula is:
∫ u d v = u v − ∫ v d u \int u dv = uv - \int v du ∫ u d v = uv − ∫ v d u The choice of u u u d v dv d v u u u d v dv d v d u du d u v v v
Example I
∫ x e − x d x \int x e^{-x} \, dx ∫ x e − x d x Let u = x u = x u = x d u = d x du = dx d u = d x d v = e − x d x dv = e^{-x} dx d v = e − x d x v = − e − x v = -e^{-x} v = − e − x
By the integration by parts formula ∫ u d v = u v − ∫ v d u \int u \, dv = uv - \int v \, du ∫ u d v = uv − ∫ v d u
∫ x e − x d x = u v − ∫ v d u \int x e^{-x} \, dx = uv - \int v \, du ∫ x e − x d x = uv − ∫ v d u ∫ x e − x d x = − x e − x − ∫ − e − x d x \int x e^{-x} \, dx = -x e^{-x} - \int -e^{-x} \, dx ∫ x e − x d x = − x e − x − ∫ − e − x d x Integrating − e − x -e^{-x} − e − x e − x e^{-x} e − x
∫ x e − x d x = − x e − x + e − x + C \int x e^{-x} \, dx = -x e^{-x} + e^{-x} + C ∫ x e − x d x = − x e − x + e − x + C Where C C C
Example II
∫ 3 5 ln ( x ) d x \int_{3}^{5} \ln(x) \, dx ∫ 3 5 ln ( x ) d x Let u = ln ( x ) u = \ln(x) u = ln ( x ) d u = 1 x d x du = \frac{1}{x}dx d u = x 1 d x d v = d x dv = dx d v = d x v = x v = x v = x
Using the integration by parts formula ∫ a b u d v = u v ∣ a b − ∫ a b v d u \int_{a}^{b} u \, dv = uv \bigg|_{a}^{b} - \int_{a}^{b} v \, du ∫ a b u d v = uv a b − ∫ a b v d u
∫ 3 5 ln ( x ) d x = x ln ( x ) ∣ 3 5 − ∫ 3 5 x 1 x d x \int_{3}^{5} \ln(x) \, dx = x\ln(x) \bigg|_{3}^{5} - \int_{3}^{5} x \, \frac{1}{x} \, dx ∫ 3 5 ln ( x ) d x = x ln ( x ) 3 5 − ∫ 3 5 x x 1 d x Simplifying the integral ∫ 3 5 x 1 x d x \int_{3}^{5} x \, \frac{1}{x} \, dx ∫ 3 5 x x 1 d x ∫ 3 5 d x \int_{3}^{5} dx ∫ 3 5 d x
∫ 3 5 ln ( x ) d x = x ln ( x ) ∣ 3 5 − ∫ 3 5 d x \int_{3}^{5} \ln(x) \, dx = x\ln(x) \bigg|_{3}^{5} - \int_{3}^{5} dx ∫ 3 5 ln ( x ) d x = x ln ( x ) 3 5 − ∫ 3 5 d x ∫ 3 5 ln ( x ) d x = x ln ( x ) ∣ 3 5 − x ∣ 3 5 \int_{3}^{5} \ln(x) \, dx = x\ln(x) \bigg|_{3}^{5} - x \bigg|_{3}^{5} ∫ 3 5 ln ( x ) d x = x ln ( x ) 3 5 − x 3 5 Subtracting 5 − 3 5 - 3 5 − 3 5 ln ( 5 ) − 3 ln ( 3 ) 5\ln(5) - 3\ln(3) 5 ln ( 5 ) − 3 ln ( 3 )
∫ 3 5 ln ( x ) d x = 5 ln ( 5 ) − 3 ln ( 3 ) − ( 5 − 3 ) \int_{3}^{5} \ln(x) \, dx = 5\ln(5) - 3\ln(3) - (5 - 3) ∫ 3 5 ln ( x ) d x = 5 ln ( 5 ) − 3 ln ( 3 ) − ( 5 − 3 ) ∫ 3 5 ln ( x ) d x = 5 ln ( 5 ) − 3 ln ( 3 ) − 2 \int_{3}^{5} \ln(x) \, dx = 5\ln(5) - 3\ln(3) - 2 ∫ 3 5 ln ( x ) d x = 5 ln ( 5 ) − 3 ln ( 3 ) − 2 Trigonometric Substitutions in Integration
Case 1: a 2 − b 2 x 2 \sqrt{a^2 - b^2x^2} a 2 − b 2 x 2
Substitution: x = a b sin ( θ ) x = \frac{a}{b} \sin(\theta) x = b a sin ( θ ) Identity: cos 2 ( θ ) = 1 − sin 2 ( θ ) \cos^2(\theta) = 1 - \sin^2(\theta) cos 2 ( θ ) = 1 − sin 2 ( θ )
Case 2: b 2 x 2 − a 2 \sqrt{b^2x^2 - a^2} b 2 x 2 − a 2
Substitution: x = a b sec ( θ ) x = \frac{a}{b} \sec(\theta) x = b a sec ( θ ) Identity: tan 2 ( θ ) = sec 2 ( θ ) − 1 \tan^2(\theta) = \sec^2(\theta) - 1 tan 2 ( θ ) = sec 2 ( θ ) − 1
Case 3: a 2 + b 2 x 2 \sqrt{a^2 + b^2x^2} a 2 + b 2 x 2
Substitution: x = a b tan ( θ ) x = \frac{a}{b} \tan(\theta) x = b a tan ( θ ) Identity: sec 2 ( θ ) = 1 + tan 2 ( θ ) \sec^2(\theta) = 1 + \tan^2(\theta) sec 2 ( θ ) = 1 + tan 2 ( θ )
Example
∫ 16 x 2 4 − 9 x 2 d x \int \frac{16}{x^2 \sqrt{4-9x^2}} \, dx ∫ x 2 4 − 9 x 2 16 d x Let x = 2 3 sin ( θ ) x = \frac{2}{3} \sin(\theta) x = 3 2 sin ( θ ) d x = 2 3 cos ( θ ) d θ dx = \frac{2}{3}\cos(\theta)d\theta d x = 3 2 cos ( θ ) d θ
4 − 9 x 2 = 4 − 4 sin 2 ( θ ) = 4 cos 2 ( θ ) = 2 ∣ cos ( θ ) ∣ \sqrt{4-9x^2}=\sqrt{4-4\sin^2(\theta)}=\sqrt{4\cos^2(\theta)}= 2 \mid \cos(\theta) \mid 4 − 9 x 2 = 4 − 4 sin 2 ( θ ) = 4 cos 2 ( θ ) = 2 ∣ cos ( θ ) ∣ In this case we have 4 − 9 x 2 = 2 cos ( θ ) \sqrt{4-9x^2} = 2\cos(\theta) 4 − 9 x 2 = 2 cos ( θ )
∫ 16 x 2 4 − 9 x 2 d x = ∫ 16 ( 2 3 sin ( θ ) ) 2 ⋅ 2 cos ( θ ) ⋅ 2 3 cos ( θ ) d θ = ∫ 12 sin 2 ( θ ) d θ \int \frac{16}{x^2 \sqrt{4-9x^2}} \, dx = \int \frac{16}{\left(\frac{2}{3}\sin(\theta)\right)^2 \cdot 2\cos(\theta)} \cdot \frac{2}{3}\cos(\theta) \, d\theta = \int \frac{12}{\sin^2(\theta)} \, d\theta ∫ x 2 4 − 9 x 2 16 d x = ∫ ( 3 2 sin ( θ ) ) 2 ⋅ 2 cos ( θ ) 16 ⋅ 3 2 cos ( θ ) d θ = ∫ sin 2 ( θ ) 12 d θ ∫ 16 ( 2 3 sin ( θ ) ) 2 ⋅ 2 cos ( θ ) ⋅ 2 3 cos ( θ ) d θ = ∫ 12 csc 2 ( θ ) d θ \int \frac{16}{\left(\frac{2}{3}\sin(\theta)\right)^2 \cdot 2\cos(\theta)} \cdot \frac{2}{3}\cos(\theta) \, d\theta = \int 12 \csc^2(\theta) \, d\theta ∫ ( 3 2 sin ( θ ) ) 2 ⋅ 2 cos ( θ ) 16 ⋅ 3 2 cos ( θ ) d θ = ∫ 12 csc 2 ( θ ) d θ This simplification uses the identity sin 2 ( θ ) + cos 2 ( θ ) = 1 \sin^2(\theta) + \cos^2(\theta) = 1 sin 2 ( θ ) + cos 2 ( θ ) = 1 csc ( θ ) = 1 sin ( θ ) \csc(\theta) = \frac{1}{\sin(\theta)} csc ( θ ) = s i n ( θ ) 1
The antiderivative of csc 2 ( θ ) \csc^2(\theta) csc 2 ( θ ) − cot ( θ ) - \cot(\theta) − cot ( θ )
∫ 12 csc 2 ( θ ) d θ = − 12 cot ( θ ) + C \int 12 \csc^2(\theta) \, d\theta = -12 \cot(\theta) + C ∫ 12 csc 2 ( θ ) d θ = − 12 cot ( θ ) + C Using Right Triangle Trigonometry
To convert back to x x x sin ( θ ) = 3 x 2 \sin(\theta) = \frac{3x}{2} sin ( θ ) = 2 3 x
The triangle formed by the substitution suggests that:
cot ( θ ) = 4 − 9 x 2 3 x \cot(\theta) = \frac{\sqrt{4 - 9x^2}}{3x} cot ( θ ) = 3 x 4 − 9 x 2 Final Integral in terms of X
Substituting cot ( θ ) \cot(\theta) cot ( θ )
− 12 cot ( θ ) + C = − 4 4 − 9 x 2 1 x + C -12 \cot(\theta) + C = -4\sqrt{4 - 9x^2} \frac{1}{x} + C − 12 cot ( θ ) + C = − 4 4 − 9 x 2 x 1 + C Thus, the final antiderivative in terms of x x x
∫ 16 x 2 4 − 9 x 2 d x = − 4 4 − 9 x 2 x + C \int \frac{16}{x^2 \sqrt{4 - 9x^2}} \, dx = -4 \frac{\sqrt{4 - 9x^2}}{x} + C ∫ x 2 4 − 9 x 2 16 d x = − 4 x 4 − 9 x 2 + C Partial Fraction Decomposition for Integration
Concept
Partial fractions is a technique used to simplify the integration of rational expressions, P ( x ) Q ( x ) \frac{P(x)}{Q(x)} Q ( x ) P ( x ) P ( x ) P(x) P ( x ) Q ( x ) Q(x) Q ( x )
Steps for Decomposition
Factor the denominator Q ( x ) Q(x) Q ( x )
Write down a partial fraction for each factor in Q ( x ) Q(x) Q ( x )
Solve for the constants by clearing the denominators and equating coefficients for corresponding powers of x x x
Table for Decomposition
Factor of Q ( x ) Q(x) Q ( x ) Term in Partial Fraction Decomposition (P.F.D) a x + b ax + b a x + b A a x + b \frac{A}{ax + b} a x + b A a x 2 + b x + c ax^2 + bx + c a x 2 + b x + c A x + B a x 2 + b x + c \frac{Ax + B}{ax^2 + bx + c} a x 2 + b x + c A x + B ( a x + b ) k (ax + b)^k ( a x + b ) k A 1 a x + b + A 2 ( a x + b ) 2 + ⋯ + A k ( a x + b ) k \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_k}{(ax + b)^k} a x + b A 1 + ( a x + b ) 2 A 2 + ⋯ + ( a x + b ) k A k ( a x 2 + b x + c ) k (ax^2 + bx + c)^k ( a x 2 + b x + c ) k A 1 x + B 1 a x 2 + b x + c + ⋯ + A k x + B k ( a x 2 + b x + c ) k \frac{A_1x + B_1}{ax^2 + bx + c} + \cdots + \frac{A_kx + B_k}{(ax^2 + bx + c)^k} a x 2 + b x + c A 1 x + B 1 + ⋯ + ( a x 2 + b x + c ) k A k x + B k
Example of Integration Using Partial Fraction
∫ 7 x 2 + 13 x ( x − 1 ) ( x 2 + 4 ) d x \int \frac{7x^2 + 13x}{(x - 1)(x^2 + 4)} \, dx ∫ ( x − 1 ) ( x 2 + 4 ) 7 x 2 + 13 x d x Partial Fraction Decomposition
To decompose the function, assume it can be written as the sum of fractions:
7 x 2 + 13 x ( x − 1 ) ( x 2 + 4 ) = A x − 1 + B x + C x 2 + 4 \frac{7x^2 + 13x}{(x - 1)(x^2 + 4)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 4} ( x − 1 ) ( x 2 + 4 ) 7 x 2 + 13 x = x − 1 A + x 2 + 4 B x + C Where A A A B B B C C C
Finding Constants
Multiply both sides by the common denominator ( x − 1 ) ( x 2 + 4 ) (x - 1)(x^2 + 4) ( x − 1 ) ( x 2 + 4 )
7 x 2 + 13 x = A ( x 2 + 4 ) + ( B x + C ) ( x − 1 ) 7x^2 + 13x = A(x^2 + 4) + (Bx + C)(x - 1) 7 x 2 + 13 x = A ( x 2 + 4 ) + ( B x + C ) ( x − 1 ) Expanding the right side and collecting like terms gives:
7 x 2 + 13 x = ( A + B ) x 2 + ( C − B ) x + 4 A − C 7x^2 + 13x = (A + B)x^2 + (C - B)x + 4A - C 7 x 2 + 13 x = ( A + B ) x 2 + ( C − B ) x + 4 A − C Setting Coefficients Equal
Match the coefficients of corresponding powers of x x x
x 2 : A + B = 7 x : C − B = 13 Constant : 4 A − C = 0 \begin{align*}
x^2: & \quad A + B = 7 \\
x: & \quad C - B = 13 \\
\text{Constant}: & \quad 4A - C = 0
\end{align*} x 2 : x : Constant : A + B = 7 C − B = 13 4 A − C = 0 Solving the System of Equations
Solve this system of linear equations to find the values of A A A B B B C C C
A = 4 B = 3 C = 16 \begin{align*}
A & = 4 \\
B & = 3 \\
C & = 16
\end{align*} A B C = 4 = 3 = 16 Rewriting the Integral
Substitute the values of A A A B B B C C C
∫ 4 x − 1 + 3 x + 16 x 2 + 4 d x \int \frac{4}{x - 1} + \frac{3x + 16}{x^2 + 4} \, dx ∫ x − 1 4 + x 2 + 4 3 x + 16 d x Integrating Each Term
For 4 x − 1 \frac{4}{x - 1} x − 1 4 4 ln ∣ x − 1 ∣ 4 \ln|x - 1| 4 ln ∣ x − 1∣
For 3 x x 2 + 4 \frac{3x}{x^2 + 4} x 2 + 4 3 x u = x 2 + 4 u = x^2 + 4 u = x 2 + 4 3 2 ln ∣ x 2 + 4 ∣ \frac{3}{2} \ln|x^2 + 4| 2 3 ln ∣ x 2 + 4∣
For 16 x 2 + 4 \frac{16}{x^2 + 4} x 2 + 4 16 8 tan − 1 ( x 2 ) 8\tan^{-1}\left(\frac{x}{2}\right) 8 tan − 1 ( 2 x ) d d x [ tan − 1 ( x 2 ) ] = 2 x 2 + 4 \frac{d}{dx}\left[\tan^{-1}\left(\frac{x}{2}\right)\right] = \frac{2}{x^2 + 4} d x d [ tan − 1 ( 2 x ) ] = x 2 + 4 2
Final Answer
Combine all antiderivatives and add the constant of integration to get the final solution:
∫ 7 x 2 + 13 x ( x − 1 ) ( x 2 + 4 ) d x = 4 ln ∣ x − 1 ∣ + 3 2 ln ∣ x 2 + 4 ∣ + 8 tan − 1 ( x 2 ) + C \int \frac{7x^2 + 13x}{(x - 1)(x^2 + 4)} \, dx = 4 \ln|x - 1| + \frac{3}{2} \ln|x^2 + 4| + 8\tan^{-1}\left(\frac{x}{2}\right) + C ∫ ( x − 1 ) ( x 2 + 4 ) 7 x 2 + 13 x d x = 4 ln ∣ x − 1∣ + 2 3 ln ∣ x 2 + 4∣ + 8 tan − 1 ( 2 x ) + C