Integration by parts: logarithms, powers and a returning integral

Choose u=ln x, dv/dx=x², so du/dx=1/x and v=x³/3. Integration by parts gives $(x^3/3)\ln x-(1/3)\int x^2\,dx$; x>0.

The integral is $\dfrac{x^3}3\ln x-\dfrac{x^3}9+C$.

Choose u=ln(x²+1), dv/dx=1, so du/dx=2x/(x²+1), v=x. Parts gives $x\ln(x^2+1)-2\int x^2/(x^2+1)\,dx$. Rewrite x²/(x²+1)=1−1/(x²+1) to integrate the remainder.

The integral is $x\ln(x^2+1)-2x+2\tan^{-1}x+C$.

Choose u=arctan(x²), dv/dx=x, so du/dx=2x/(1+x⁴), v=x²/2. Parts gives $(x^2/2)\tan^{-1}(x^2)-\int x^3/(1+x^4)\,dx$. Insert the factor 4 to obtain the logarithm. The source removes absolute values because 1+x⁴>0.

The integral is $\dfrac{x^2}2\tan^{-1}(x^2)-\dfrac14\ln(1+x^4)+C$.

First choose u=x², dv/dx=eˣ, giving du/dx=2x, v=eˣ. Parts yields $x^2e^x-2\int xe^x\,dx$. Apply parts again with u′=x, dv′/dx=eˣ, du′/dx=1, v′=eˣ. Then $\int xe^x\,dx=xe^x-\int e^x\,dx$. Substitution gives the stated result.

The integral is $x^2e^x-2xe^x+2e^x+C$.

Let I be the required integral. Choose u=cos x and dv/dx=eˣ, so du/dx=−sin x, v=eˣ. Then $I=e^x\cos x+\int e^x\sin x\,dx$. Apply parts to the remaining integral with u′=sin x, dv′/dx=eˣ, du′/dx=cos x, v′=eˣ. Thus $I=e^x\cos x+e^x\sin x-I$. Bring I to the left: $2I=e^x\cos x+e^x\sin x+C^\prime$.

The integral is $\dfrac12e^x\cos x+\dfrac12e^x\sin x+C$.

Use $2\sin\theta\cos\theta=\sin2\theta$ to make the integral $(1/2)\int\theta\sin2\theta\,d\theta$. Choose u=θ and dv/dθ=sin2θ, with du/dθ=1 and v=−cos2θ/2. Parts gives $-\theta\cos2\theta/4+(1/4)\int\cos2\theta\,d\theta$, leading to the sine term.

The integral is $-\dfrac{\theta\cos2\theta}4+\dfrac18\sin2\theta+C$.