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  1. High Level Math
  2. Integration
  3. Indefinite integral

Integration by parts

Integration by parts is a method to find integrals of products.

Product Rule→Integration by Partsddx[f(x)g(x)]=f′(x)g(x)+f(x)g′(x)f(x)g(x)=∫f′(x)g(x)⋅dx+∫f(x)g′(x)⋅dxf(x)g(x)−∫f′(x)g(x)⋅dx=∫f(x)g′(x)⋅dx⇓∫f(x)g′(x)⋅dx=f(x)g(x)−∫f′(x)g(x)⋅dx\begin{align*} \text{Product Rule} &\rightarrow \text{Integration by Parts} \\ \\ \frac{d}{dx}[f(x)g(x)] &= f^\prime(x)g(x) + f(x)g^\prime(x) \\ \\ f(x)g(x) &= \int f^\prime(x)g(x) \cdot dx + \int f(x)g^\prime(x) \cdot dx \\ \\ f(x)g(x) - \int f^\prime(x)g(x) \cdot dx &= \int f(x)g^\prime(x) \cdot dx \\ \\ &\Downarrow \\ \\ \int f(x)g^\prime(x) \cdot dx &= f(x)g(x) - \int f^\prime(x)g(x) \cdot dx \end{align*}Product Ruledxd​[f(x)g(x)]f(x)g(x)f(x)g(x)−∫f′(x)g(x)⋅dx∫f(x)g′(x)⋅dx​→Integration by Parts=f′(x)g(x)+f(x)g′(x)=∫f′(x)g(x)⋅dx+∫f(x)g′(x)⋅dx=∫f(x)g′(x)⋅dx⇓=f(x)g(x)−∫f′(x)g(x)⋅dx​
∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)⋅dx∫xcos⁡x⋅dx=xsin⁡x−∫1sin⁡x⋅dx=xsin⁡x−(−cos⁡x+C)=xsin⁡x+cos⁡x−C=xsin⁡x+cos⁡x+C\begin{align*} \int f(x)g^\prime(x)dx &= f(x)g(x) - \int f^\prime(x)g(x) \cdot dx \\ \\ \int x \cos{x} \cdot dx &= x \sin{x} - \int 1 \sin{x} \cdot dx \\ \\ &= x \sin{x} - (-\cos{x} + C) \\ \\ &= x \sin{x} + \cos{x} - C \\ \\ &= x \sin{x} + \cos{x} + C \end{align*}∫f(x)g′(x)dx∫xcosx⋅dx​=f(x)g(x)−∫f′(x)g(x)⋅dx=xsinx−∫1sinx⋅dx=xsinx−(−cosx+C)=xsinx+cosx−C=xsinx+cosx+C​
∫f(x)g′(x)⋅dx=f(x)g(x)−∫f′(x)g(x)⋅dx∫(ln⁡x)dx=∫(ln⁡x)⋅1dx=ln⁡x⋅x−∫1x⋅xdx=ln⁡x⋅x−x+C\begin{align*} & \int f(x)g^\prime(x) \cdot dx = f(x)g(x) - \int f^\prime(x)g(x) \cdot dx \\ \\ \int (\ln{x}) dx = \int (\ln{x}) \cdot 1 dx &= \ln{x} \cdot x - \int \frac{1}{x} \cdot x dx \\ \\ &= \ln{x} \cdot x - x + C \end{align*}∫(lnx)dx=∫(lnx)⋅1dx​∫f(x)g′(x)⋅dx=f(x)g(x)−∫f′(x)g(x)⋅dx=lnx⋅x−∫x1​⋅xdx=lnx⋅x−x+C​
∫f(x)g′(x)⋅dx=f(x)g(x)−∫f′(x)g(x)⋅dx∫x2exdx=x2ex−∫2xexdx∫x2exdx=x2ex−2∫xexdx‾∫xexdx‾=xex−∫1exdx=xex−ex∫x2exdx=x2ex−2(xex−ex‾)=x2ex−2xex+2ex\begin{align*} \int f(x)g^\prime(x) \cdot dx &= f(x)g(x) - \int f^\prime(x)g(x) \cdot dx \\ \\ \\ \int x^2 e^x dx &= x^2 e^x - \int 2x e^x dx \\ \\ \int x^2 e^x dx &= x^2 e^x - 2\underline{\int x e^x dx} \\ \\ \\ \underline{\int x e^x dx} &= x e^x - \int 1 e^x dx \\ \\ &= x e^x - e^x \\ \\ \\ \int x^2 e^x dx &= x^2 e^x - 2(\underline{x e^x - e^x}) \\ \\ &= x^2 e^x - 2x e^x + 2e^x \end{align*}∫f(x)g′(x)⋅dx∫x2exdx∫x2exdx∫xexdx​∫x2exdx​=f(x)g(x)−∫f′(x)g(x)⋅dx=x2ex−∫2xexdx=x2ex−2∫xexdx​=xex−∫1exdx=xex−ex=x2ex−2(xex−ex​)=x2ex−2xex+2ex​
∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx∫excos⁡xdx=exsin⁡x−∫exsin⁡xdx‾∫exsin⁡xdx‾=ex(−cos⁡x)−∫ex(−cos⁡x)dx=−excos⁡x+∫excos⁡xdx∫excos⁡xdx=exsin⁡x−(−excos⁡x+∫excos⁡xdx‾)=exsin⁡x+excos⁡x−∫excos⁡xdx2∫excos⁡xdx=exsin⁡x+excos⁡x∫excos⁡xdx=exsin⁡x+excos⁡x2+C\begin{align*} \int f(x)g^\prime(x) dx &= f(x)g(x) - \int f^\prime(x)g(x) dx \\ \\ \\ \int e^x \cos{x} dx &= e^x \sin{x} - \underline{\int e^x \sin{x} dx} \\ \\ \\ \underline{\int e^x \sin{x} dx} &= e^x (-\cos{x}) - \int e^x (-\cos{x}) dx \\ \\ &= - e^x \cos{x} + \int e^x \cos{x} dx \\ \\ \\ \int e^x \cos{x} dx &= e^x \sin{x} - (\underline{- e^x \cos{x} + \int e^x \cos{x} dx}) \\ \\ &= e^x \sin{x} + e^x \cos{x} - \int e^x \cos{x} dx \\ \\ 2 \int e^x \cos{x} dx &= e^x \sin{x} + e^x \cos{x} \\ \\ \int e^x \cos{x} dx &= \frac{e^x \sin{x} + e^x \cos{x}}{2} + C \end{align*}∫f(x)g′(x)dx∫excosxdx∫exsinxdx​∫excosxdx2∫excosxdx∫excosxdx​=f(x)g(x)−∫f′(x)g(x)dx=exsinx−∫exsinxdx​=ex(−cosx)−∫ex(−cosx)dx=−excosx+∫excosxdx=exsinx−(−excosx+∫excosxdx​)=exsinx+excosx−∫excosxdx=exsinx+excosx=2exsinx+excosx​+C​

Integration by parts is a method to find integrals of products.

Product Rule→Integration by Partsddx[f(x)g(x)]=f′(x)g(x)+f(x)g′(x)f(x)g(x)=∫f′(x)g(x)⋅dx+∫f(x)g′(x)⋅dxf(x)g(x)−∫f′(x)g(x)⋅dx=∫f(x)g′(x)⋅dx⇓∫f(x)g′(x)⋅dx=f(x)g(x)−∫f′(x)g(x)⋅dx\begin{align*} \text{Product Rule} &\rightarrow \text{Integration by Parts} \\ \\ \frac{d}{dx}[f(x)g(x)] &= f^\prime(x)g(x) + f(x)g^\prime(x) \\ \\ f(x)g(x) &= \int f^\prime(x)g(x) \cdot dx + \int f(x)g^\prime(x) \cdot dx \\ \\ f(x)g(x) - \int f^\prime(x)g(x) \cdot dx &= \int f(x)g^\prime(x) \cdot dx \\ \\ &\Downarrow \\ \\ \int f(x)g^\prime(x) \cdot dx &= f(x)g(x) - \int f^\prime(x)g(x) \cdot dx \end{align*}Product Ruledxd​[f(x)g(x)]f(x)g(x)f(x)g(x)−∫f′(x)g(x)⋅dx∫f(x)g′(x)⋅dx​→Integration by Parts=f′(x)g(x)+f(x)g′(x)=∫f′(x)g(x)⋅dx+∫f(x)g′(x)⋅dx=∫f(x)g′(x)⋅dx⇓=f(x)g(x)−∫f′(x)g(x)⋅dx​
∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)⋅dx∫xcos⁡x⋅dx=xsin⁡x−∫1sin⁡x⋅dx=xsin⁡x−(−cos⁡x+C)=xsin⁡x+cos⁡x−C=xsin⁡x+cos⁡x+C\begin{align*} \int f(x)g^\prime(x)dx &= f(x)g(x) - \int f^\prime(x)g(x) \cdot dx \\ \\ \int x \cos{x} \cdot dx &= x \sin{x} - \int 1 \sin{x} \cdot dx \\ \\ &= x \sin{x} - (-\cos{x} + C) \\ \\ &= x \sin{x} + \cos{x} - C \\ \\ &= x \sin{x} + \cos{x} + C \end{align*}∫f(x)g′(x)dx∫xcosx⋅dx​=f(x)g(x)−∫f′(x)g(x)⋅dx=xsinx−∫1sinx⋅dx=xsinx−(−cosx+C)=xsinx+cosx−C=xsinx+cosx+C​
∫f(x)g′(x)⋅dx=f(x)g(x)−∫f′(x)g(x)⋅dx∫(ln⁡x)dx=∫(ln⁡x)⋅1dx=ln⁡x⋅x−∫1x⋅xdx=ln⁡x⋅x−x+C\begin{align*} & \int f(x)g^\prime(x) \cdot dx = f(x)g(x) - \int f^\prime(x)g(x) \cdot dx \\ \\ \int (\ln{x}) dx = \int (\ln{x}) \cdot 1 dx &= \ln{x} \cdot x - \int \frac{1}{x} \cdot x dx \\ \\ &= \ln{x} \cdot x - x + C \end{align*}∫(lnx)dx=∫(lnx)⋅1dx​∫f(x)g′(x)⋅dx=f(x)g(x)−∫f′(x)g(x)⋅dx=lnx⋅x−∫x1​⋅xdx=lnx⋅x−x+C​
∫f(x)g′(x)⋅dx=f(x)g(x)−∫f′(x)g(x)⋅dx∫x2exdx=x2ex−∫2xexdx∫x2exdx=x2ex−2∫xexdx‾∫xexdx‾=xex−∫1exdx=xex−ex∫x2exdx=x2ex−2(xex−ex‾)=x2ex−2xex+2ex\begin{align*} \int f(x)g^\prime(x) \cdot dx &= f(x)g(x) - \int f^\prime(x)g(x) \cdot dx \\ \\ \\ \int x^2 e^x dx &= x^2 e^x - \int 2x e^x dx \\ \\ \int x^2 e^x dx &= x^2 e^x - 2\underline{\int x e^x dx} \\ \\ \\ \underline{\int x e^x dx} &= x e^x - \int 1 e^x dx \\ \\ &= x e^x - e^x \\ \\ \\ \int x^2 e^x dx &= x^2 e^x - 2(\underline{x e^x - e^x}) \\ \\ &= x^2 e^x - 2x e^x + 2e^x \end{align*}∫f(x)g′(x)⋅dx∫x2exdx∫x2exdx∫xexdx​∫x2exdx​=f(x)g(x)−∫f′(x)g(x)⋅dx=x2ex−∫2xexdx=x2ex−2∫xexdx​=xex−∫1exdx=xex−ex=x2ex−2(xex−ex​)=x2ex−2xex+2ex​
∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx∫excos⁡xdx=exsin⁡x−∫exsin⁡xdx‾∫exsin⁡xdx‾=ex(−cos⁡x)−∫ex(−cos⁡x)dx=−excos⁡x+∫excos⁡xdx∫excos⁡xdx=exsin⁡x−(−excos⁡x+∫excos⁡xdx‾)=exsin⁡x+excos⁡x−∫excos⁡xdx2∫excos⁡xdx=exsin⁡x+excos⁡x∫excos⁡xdx=exsin⁡x+excos⁡x2+C\begin{align*} \int f(x)g^\prime(x) dx &= f(x)g(x) - \int f^\prime(x)g(x) dx \\ \\ \\ \int e^x \cos{x} dx &= e^x \sin{x} - \underline{\int e^x \sin{x} dx} \\ \\ \\ \underline{\int e^x \sin{x} dx} &= e^x (-\cos{x}) - \int e^x (-\cos{x}) dx \\ \\ &= - e^x \cos{x} + \int e^x \cos{x} dx \\ \\ \\ \int e^x \cos{x} dx &= e^x \sin{x} - (\underline{- e^x \cos{x} + \int e^x \cos{x} dx}) \\ \\ &= e^x \sin{x} + e^x \cos{x} - \int e^x \cos{x} dx \\ \\ 2 \int e^x \cos{x} dx &= e^x \sin{x} + e^x \cos{x} \\ \\ \int e^x \cos{x} dx &= \frac{e^x \sin{x} + e^x \cos{x}}{2} + C \end{align*}∫f(x)g′(x)dx∫excosxdx∫exsinxdx​∫excosxdx2∫excosxdx∫excosxdx​=f(x)g(x)−∫f′(x)g(x)dx=exsinx−∫exsinxdx​=ex(−cosx)−∫ex(−cosx)dx=−excosx+∫excosxdx=exsinx−(−excosx+∫excosxdx​)=exsinx+excosx−∫excosxdx=exsinx+excosx=2exsinx+excosx​+C​

The priciple of Integration by parts is: whose derivative f′f^\primef′ simpler, who's gonna be f(x)f(x)f(x)

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Last updated 6 years ago

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