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Integration by Parts

The Product Rule states that if ff and gg are differentiable functions, then
ddx[f(x)g(x)]=f(x)g′(x)+g(x)f′(x).\begin{equation*}\frac{\mathrm{d}}{\mathrm{d}x}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x).\end{equation*}
In the notation for indefinite integrals this equation becomes after rearrangement
∫f(x)g′(x) dx=f(x)g(x)−∫g(x)f′(x) dx.\begin{equation*}\int f(x)g'(x)\,\mathrm{d}x=f(x)g(x)-\int g(x)f'(x)\,\mathrm{d}x.\end{equation*}
It is easier to remember in the following notation.
INTEGRATION BY PARTS
Let u=f(x)u=f(x) and v=g(x)v=g(x). Then the differentials are du=f′(x) dx\mathrm{d}u=f'(x)\,\mathrm{d}x and dv=g′(x) dx\mathrm{d}v=g'(x)\,\mathrm{d}x, the formula for integration by parts becomes
∫u dv=uv−∫v du.\begin{equation*}\int u\,\mathrm{d}v=uv-\int v\,\mathrm{d}u.\end{equation*}

Examples

∫ln⁡x dx\int \ln x\,\mathrm{d}x

Let's evaluate ∫ln⁡x dx\int \ln x\,\mathrm{d}x for x>0x>0.
Define
u=ln⁡x,dv=dx.\begin{equation*}u=\ln x,\qquad \mathrm{d}v=\mathrm{d}x.\end{equation*}
Then
du=1x dx,v=x.\begin{equation*}\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x,\qquad v=x.\end{equation*}
Integrating by parts, we get
∫ln⁡x dx=xln⁡x−∫x⋅1x dx=xln⁡x−∫dx=xln⁡x−x+C.\begin{align*}\int \ln x\,\mathrm{d}x &= x\ln x-\int x\cdot\frac{1}{x}\,\mathrm{d}x \\&= x\ln x-\int \mathrm{d}x \\&= x\ln x-x+C.\end{align*}

∫exsin⁡x dx\int e^x\sin x\,\mathrm{d}x

Define
u=ex,dv=sin⁡x dx.\begin{equation*}u=e^x,\qquad \mathrm{d}v=\sin x\,\mathrm{d}x.\end{equation*}
Then
du=ex dx,v=−cos⁡x.\begin{equation*}\mathrm{d}u=e^x\,\mathrm{d}x,\qquad v=-\cos x.\end{equation*}
Integrating by parts, we get
∫exsin⁡x dx=−excos⁡x+∫excos⁡x dx.\begin{equation*}\int e^x\sin x\,\mathrm{d}x=-e^x\cos x+\int e^x\cos x\,\mathrm{d}x.\end{equation*}
The remaining integral is no simpler, but we can integrate by parts again. Keeping u=exu=e^x, let
u=ex,dv=cos⁡x dx.\begin{equation*}u=e^x,\qquad \mathrm{d}v=\cos x\,\mathrm{d}x.\end{equation*}
Then
du=ex dx,v=sin⁡x,\begin{equation*}\mathrm{d}u=e^x\,\mathrm{d}x,\qquad v=\sin x,\end{equation*}
so
∫excos⁡x dx=exsin⁡x−∫exsin⁡x dx.\begin{equation*}\int e^x\cos x\,\mathrm{d}x=e^x\sin x-\int e^x\sin x\,\mathrm{d}x.\end{equation*}
The original integral appears again. Substituting this expression into the first equation gives
∫exsin⁡x dx=−excos⁡x+exsin⁡x−∫exsin⁡x dx.\begin{equation*}\int e^x\sin x\,\mathrm{d}x=-e^x\cos x+e^x\sin x-\int e^x\sin x\,\mathrm{d}x.\end{equation*}
After rearrangement, we obtain
∫exsin⁡x dx=12ex(sin⁡x−cos⁡x)+C.\begin{equation*}\int e^x\sin x\,\mathrm{d}x=\frac{1}{2}e^x(\sin x-\cos x)+C.\end{equation*}

References

  • [StewartEtAl2021]Stewart, James, Daniel K. Clegg, and Saleem Watson. Calculus: Early Transcendentals, Metric Edition. Ninth edition. Cengage, 2021.
  • [TheMathFlow2026]The Math Flow. ``The Geometry of Integration by Parts.'' X, 2026. Video.