Question

Evaluate the indefinite integral as a power series. x2 ln(1 + x) dx

Answer 1

$\ln(1+x)= \int { \frac{1}{1+x} } \, dx \\ \\ =\int { \frac{1}{1-(-x)} } \, dx=\Sigma_{k=0}^{\infty}(-x)^k \\ \\ =\Sigma_{k=0}^{\infty}(-1)^kx^k \\ \\ \Rightarrow x^2\ln(1+x)=x^2\Sigma_{k=0}^{\infty}(-1)^kx^k \\ \\ =\Sigma_{k=0}^{\infty}(-1)^kx^{k+2} \\ \\ \Rightarrow \int {x^2\ln(1+x)} \, dx =\int {\Sigma_{k=0}^{\infty}(-1)^kx^{k+2}} \, dx \\ \\ =C+\Sigma_{k=0}^{\infty}(-1)^k\frac{x^{k+3}}{k+3}$

When x = 0. we have

$\int {0^2\ln(1+0)}=C+\Sigma_{k=0}^{\infty}(-1)^k\frac{0^{k+3}}{0+3} \\ \\ \Rightarrow C=0 \\ \\ \therefore\int {x^2\ln(1+x)}=\Sigma_{k=0}^{\infty}(-1)^k\left(\frac{x^{k+3}}{k+3}\right)$