How to Integrate by Parts

Indefinite Integral

Consider the integral below. We see that the integrand is a product of two functions, so it is ideal for us to integrate by parts. ∫ x e x d x {\displaystyle \int xe^{x}\mathrm {d} x} \int xe^{{x}}{\mathrm {d}}x

Recall the formula for integration by parts. This formula is very useful in the sense that it allows us to transfer the derivative from one function to another, at the cost of a minus sign and a boundary term. ∫ u d v = u v − ∫ v d u {\displaystyle \int u\mathrm {d} v=uv-\int v\mathrm {d} u} \int u{\mathrm {d}}v=uv-\int v{\mathrm {d}}u

Choose a u {\displaystyle u} u and d v , {\displaystyle \mathrm {d} v,} {\mathrm {d}}v, and find the resulting d u {\displaystyle \mathrm {d} u} {\mathrm {d}}u and v {\displaystyle v} v. We choose u = x {\displaystyle u=x} u=x because its derivative of 1 is simpler than the derivative of e x , {\displaystyle e^{x},} e^{{x}}, which is only itself. That results in d v = e x d x , {\displaystyle \mathrm {d} v=e^{x}\mathrm {d} x,} {\mathrm {d}}v=e^{{x}}{\mathrm {d}}x, whose integral is trivial. d u = d x {\displaystyle \mathrm {d} u=\mathrm {d} x} {\mathrm {d}}u={\mathrm {d}}x v = e x {\displaystyle v=e^{x}} v=e^{{x}} In general, integration of parts is a technique that aims to convert an integral into one that is simpler to integrate. If you see a product of two functions where one is a polynomial, then setting u {\displaystyle u} u to be the polynomial will most likely be a good choice. You can neglect the constant of integration when finding v , {\displaystyle v,} v, because it will drop out in the end.

Substitute these four expressions into our integral. ∫ x e x d x = x e x − ∫ e x d x {\displaystyle \int xe^{x}\mathrm {d} x=xe^{x}-\int e^{x}\mathrm {d} x} \int xe^{{x}}{\mathrm {d}}x=xe^{{x}}-\int e^{{x}}{\mathrm {d}}x The result was that our integral now consists of just one function - the exponential function. As e x {\displaystyle e^{x}} e^{{x}} is its own antiderivative with a constant, evaluating it is much easier.

Evaluate the resulting expression using any means possible. Remember to add the constant of integration, as antiderivatives are not unique. ∫ x e x d x = x e x − e x + C {\displaystyle \int xe^{x}\mathrm {d} x=xe^{x}-e^{x}+C} \int xe^{{x}}{\mathrm {d}}x=xe^{{x}}-e^{{x}}+C

Definite Integral

Consider the definite integral below. Definite integrals require evaluation at the boundaries. While the integral below looks like it has an integrand of just one function, the inverse tangent function, we can say that it is the product of inverse tangent and 1. ∫ 0 1 tan − 1 ⁡ x d x {\displaystyle \int _{0}^{1}\tan ^{-1}x\mathrm {d} x} \int _{{0}}^{{1}}\tan ^{{-1}}x{\mathrm {d}}x

Recall the integration by parts formula. ∫ a b u d v = u v | a b − ∫ a b v d u {\displaystyle \int _{a}^{b}u\mathrm {d} v=uv{\Bigg |}_{a}^{b}-\int _{a}^{b}v\mathrm {d} u} \int _{{a}}^{{b}}u{\mathrm {d}}v=uv{\Bigg |}_{{a}}^{{b}}-\int _{{a}}^{{b}}v{\mathrm {d}}u

Set u {\displaystyle u} u and d v , {\displaystyle \mathrm {d} v,} {\mathrm {d}}v, and find d u {\displaystyle \mathrm {d} u} {\mathrm {d}}u and v {\displaystyle v} v. Since the derivative of an inverse trig function is algebraic and therefore simpler, we set u = tan − 1 ⁡ x {\displaystyle u=\tan ^{-1}x} u=\tan ^{{-1}}x and d v = d x . {\displaystyle \mathrm {d} v=\mathrm {d} x.} {\mathrm {d}}v={\mathrm {d}}x. This results in d u = 1 1 + x 2 d x {\displaystyle \mathrm {d} u={\frac {1}{1+x^{2}}}\mathrm {d} x} {\mathrm {d}}u={\frac {1}{1+x^{{2}}}}{\mathrm {d}}x and v = x . {\displaystyle v=x.} v=x.

Substitute these expressions into our integral. ∫ 0 1 tan − 1 ⁡ x d x = x tan − 1 ⁡ x | 0 1 − ∫ 0 1 x 1 + x 2 d x {\displaystyle \int _{0}^{1}\tan ^{-1}x\mathrm {d} x=x\tan ^{-1}x{\Bigg |}_{0}^{1}-\int _{0}^{1}{\frac {x}{1+x^{2}}}\mathrm {d} x} \int _{{0}}^{{1}}\tan ^{{-1}}x{\mathrm {d}}x=x\tan ^{{-1}}x{\Bigg |}_{{0}}^{{1}}-\int _{{0}}^{{1}}{\frac {x}{1+x^{{2}}}}{\mathrm {d}}x

Evaluate the simplified integral using u-substitution. The numerator is proportional to the derivative of the denominator, so u-subbing is ideal. Let u = 1 + x 2 . {\displaystyle u=1+x^{2}.} u=1+x^{{2}}. Then d u = 2 x d x . {\displaystyle \mathrm {d} u=2x\mathrm {d} x.} {\mathrm {d}}u=2x{\mathrm {d}}x. Be careful in changing your boundaries. ∫ 0 1 x 1 + x 2 d x = 1 2 ∫ 1 2 1 u d u = 1 2 ln ⁡ 2 {\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {x}{1+x^{2}}}\mathrm {d} x&={\frac {1}{2}}\int _{1}^{2}{\frac {1}{u}}\mathrm {d} u\\&={\frac {1}{2}}\ln 2\end{aligned}}} {\begin{aligned}\int _{{0}}^{{1}}{\frac {x}{1+x^{{2}}}}{\mathrm {d}}x&={\frac {1}{2}}\int _{{1}}^{{2}}{\frac {1}{u}}{\mathrm {d}}u\\&={\frac {1}{2}}\ln 2\end{aligned}}

Evaluate the u v {\displaystyle uv} uv expression to complete the evaluation of the original integral. Be careful with the signs. ∫ 0 1 tan − 1 ⁡ x d x = π 4 − 1 2 ln ⁡ 2 {\displaystyle \int _{0}^{1}\tan ^{-1}x\mathrm {d} x={\frac {\pi }{4}}-{\frac {1}{2}}\ln 2} \int _{{0}}^{{1}}\tan ^{{-1}}x{\mathrm {d}}x={\frac {\pi }{4}}-{\frac {1}{2}}\ln 2

Repeated Integration by Parts

Consider the integral below. Occasionally, you may find yourself with an integral that requires multiple instances of integration by parts in order to get the desired answer. Such an integral is below. ∫ e x cos ⁡ x d x {\displaystyle \int e^{x}\cos x\mathrm {d} x} \int e^{{x}}\cos x{\mathrm {d}}x

Recall the formula for integration by parts. ∫ u d v = u v − ∫ v d u {\displaystyle \int u\mathrm {d} v=uv-\int v\mathrm {d} u} \int u{\mathrm {d}}v=uv-\int v{\mathrm {d}}u

Choose a u {\displaystyle u} u and d v , {\displaystyle \mathrm {d} v,} {\mathrm {d}}v, and find the resulting d u {\displaystyle \mathrm {d} u} {\mathrm {d}}u and v {\displaystyle v} v. As one of the functions is the exponential function, setting that as u {\displaystyle u} u will get us nowhere. Instead, let u = cos ⁡ x {\displaystyle u=\cos x} u=\cos x and d v = e x d x . {\displaystyle \mathrm {d} v=e^{x}\mathrm {d} x.} {\mathrm {d}}v=e^{{x}}{\mathrm {d}}x. What we find is that the second derivative of u {\displaystyle u} u is simply the negative of itself. That is, d 2 d x 2 cos ⁡ x = − cos ⁡ x . {\displaystyle {\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\cos x=-\cos x.} {\frac {{\mathrm {d}}^{{2}}}{{\mathrm {d}}x^{{2}}}}\cos x=-\cos x. This means that we need to integrate by parts twice to get an interesting result. d u = − sin ⁡ x {\displaystyle \mathrm {d} u=-\sin x} {\mathrm {d}}u=-\sin x v = e x {\displaystyle v=e^{x}} v=e^{{x}}

Substitute these expressions into our integral. ∫ e x cos ⁡ x d x = e x cos ⁡ x − ∫ − e x sin ⁡ x d x = e x cos ⁡ x + ∫ e x sin ⁡ x d x {\displaystyle {\begin{aligned}\int e^{x}\cos x\mathrm {d} x&=e^{x}\cos x-\int -e^{x}\sin x\mathrm {d} x\\&=e^{x}\cos x+\int e^{x}\sin x\mathrm {d} x\end{aligned}}} {\begin{aligned}\int e^{{x}}\cos x{\mathrm {d}}x&=e^{{x}}\cos x-\int -e^{{x}}\sin x{\mathrm {d}}x\\&=e^{{x}}\cos x+\int e^{{x}}\sin x{\mathrm {d}}x\end{aligned}}

Perform integration by parts on the v d u {\displaystyle v\mathrm {d} u} v{\mathrm {d}}u integral. Be careful with the signs. u = sin ⁡ x , d v = e x d x , d u = cos ⁡ x d x , v = e x {\displaystyle u=\sin x,\,\mathrm {d} v=e^{x}\mathrm {d} x,\,\mathrm {d} u=\cos {x}\mathrm {d} x,\,v=e^{x}} u=\sin x,\,{\mathrm {d}}v=e^{{x}}{\mathrm {d}}x,\,{\mathrm {d}}u=\cos {x}{\mathrm {d}}x,\,v=e^{{x}} ∫ e x sin ⁡ x d x = e x sin ⁡ x − ∫ e x cos ⁡ x d x {\displaystyle \int e^{x}\sin x\mathrm {d} x=e^{x}\sin x-\int e^{x}\cos x\mathrm {d} x} \int e^{{x}}\sin x{\mathrm {d}}x=e^{{x}}\sin x-\int e^{{x}}\cos x{\mathrm {d}}x ∫ e x cos ⁡ x d x = e x cos ⁡ x + e x sin ⁡ x − ∫ e x cos ⁡ x d x {\displaystyle \int e^{x}\cos x\mathrm {d} x=e^{x}\cos x+e^{x}\sin x-\int e^{x}\cos x\mathrm {d} x} \int e^{{x}}\cos x{\mathrm {d}}x=e^{{x}}\cos x+e^{{x}}\sin x-\int e^{{x}}\cos x{\mathrm {d}}x

Solve for the original integral. In this problem, what we have found is that by performing integration by parts twice, the original integral came up in the work. Instead of performing integration by parts endlessly, which will get us nowhere, we can solve for it instead. Don't forget the constant of integration at the very end. 2 ∫ e x cos ⁡ x d x = e x cos ⁡ x + e x sin ⁡ x ∫ e x cos ⁡ x d x = 1 2 e x cos ⁡ x + 1 2 e x sin ⁡ x + C {\displaystyle {\begin{aligned}2\int e^{x}\cos x\mathrm {d} x&=e^{x}\cos x+e^{x}\sin x\\\int e^{x}\cos x\mathrm {d} x&={\frac {1}{2}}e^{x}\cos x+{\frac {1}{2}}e^{x}\sin x+C\end{aligned}}} {\begin{aligned}2\int e^{{x}}\cos x{\mathrm {d}}x&=e^{{x}}\cos x+e^{{x}}\sin x\\\int e^{{x}}\cos x{\mathrm {d}}x&={\frac {1}{2}}e^{{x}}\cos x+{\frac {1}{2}}e^{{x}}\sin x+C\end{aligned}}

Deriving the Integration by Parts Formula

Consider the antiderivative of g ( x ) {\displaystyle g(x)} g(x). We shall call this function G ( x ) , {\displaystyle G(x),} G(x), where G {\displaystyle G} G is any function that satisfies G ′ = g . {\displaystyle G^{\prime }=g.} G^{{\prime }}=g.

Compute the derivative of f G {\displaystyle fG} fG. Since this is a product of two functions, we use the product rule. Sharp minds will intuitively see the resulting integration by parts formula as closely related to the product rule, just as u-substitution is the counterpart to the chain rule. d d x f G = f G ′ + f ′ G = f g + f ′ G {\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} x}}fG&=fG^{\prime }+f^{\prime }G\\&=fg+f^{\prime }G\end{aligned}}} {\begin{aligned}{\frac {{\mathrm {d}}}{{\mathrm {d}}x}}fG&=fG^{{\prime }}+f^{{\prime }}G\\&=fg+f^{{\prime }}G\end{aligned}}

Take the integral of both sides with respect to x {\displaystyle x} x. The above expression says that f G {\displaystyle fG} fG is the antiderivative of the right side, so we integrate both sides to recover the integral of the left side. ∫ ( f g + f ′ G ) d x = f G {\displaystyle \int (fg+f^{\prime }G)\mathrm {d} x=fG} \int (fg+f^{{\prime }}G){\mathrm {d}}x=fG

Rearrange to isolate the integral of f g {\displaystyle fg} fg. ∫ f g d x = f G − ∫ f ′ G d x {\displaystyle \int fg\mathrm {d} x=fG-\int f^{\prime }G\mathrm {d} x} \int fg{\mathrm {d}}x=fG-\int f^{{\prime }}G{\mathrm {d}}x The goal of integration by parts is seen in the expression above. We are integrating f ′ G {\displaystyle f^{\prime }G} f^{{\prime }}G instead of f g , {\displaystyle fg,} fg, and if used correctly, this results in a simpler evaluation.

Change the variables to recover the familiar compact form. We let u = f , v = G , d u = f ′ d x , d v = g d x . {\displaystyle u=f,\,v=G,\,\mathrm {d} u=f^{\prime }\mathrm {d} x,\,\mathrm {d} v=g\mathrm {d} x.} u=f,\,v=G,\,{\mathrm {d}}u=f^{{\prime }}{\mathrm {d}}x,\,{\mathrm {d}}v=g{\mathrm {d}}x. ∫ u d v = u v − ∫ v d u {\displaystyle \int u\mathrm {d} v=uv-\int v\mathrm {d} u} \int u{\mathrm {d}}v=uv-\int v{\mathrm {d}}u In general, there is no systematic process by which we can make the integral easier to evaluate. However, it is often the case that we want a u {\displaystyle u} u whose derivative is easier to manage, and a d v {\displaystyle \mathrm {d} v} {\mathrm {d}}v that can easily be integrated. For definite integrals, it is easy to show that the formula holds when writing the boundaries for all three terms, though it is important to remember that the boundaries are limits on the variable x . {\displaystyle x.} x. ∫ a b u d v = u v | a b − ∫ a b v d u {\displaystyle \int _{a}^{b}u\mathrm {d} v=uv{\Bigg |}_{a}^{b}-\int _{a}^{b}v\mathrm {d} u} \int _{{a}}^{{b}}u{\mathrm {d}}v=uv{\Bigg |}_{{a}}^{{b}}-\int _{{a}}^{{b}}v{\mathrm {d}}u