Indefinite Integrals — Question 6

Question 6 - Solution

We evaluate each term separately.

First term: $$\int x e^{-x^2}\,dx.$$ Let $u=-x^2$, so $du=-2x\,dx$: $$\int x e^{-x^2}\,dx = -\frac{1}{2}\int e^{u}\,du = -\frac{1}{2}e^{-x^2}.$$

Second term: $$\int \ln(1+x)\,dx.$$ Use integration by parts with $$u=\ln(1+x), \qquad dv=dx.$$ Then $$du=\frac{1}{1+x}\,dx, \qquad v=x.$$ Thus, $$\int \ln(1+x)\,dx = x\ln(1+x) - \int \frac{x}{1+x}\,dx.$$ Rewrite the integrand: $$\frac{x}{1+x}=1-\frac{1}{1+x}.$$ Hence, $$\int \frac{x}{1+x}\,dx = x-\ln(1+x).$$

Combining results: $$\boxed{ -\frac{1}{2}e^{-x^2} + x\ln(1+x) - x + \ln(1+x) + C }$$