Tính nguyên hàm của ((xln(x+1))/(x+1)^2)dx

* Hướng dẫn giải

Đáp án D

Đặt $\left\{\begin{array}{l}u=x.\ln \left(x+1\right)\\ \text{d}v=\frac{1}{{\left(x+1\right)}^2}\text{d}x\end{array}\right.\Rightarrow \left\{\begin{array}{l}\text{d}u=\left(\ln \left(x+1\right)+\frac{x}{x+1}\right)\text{d}x\\ v=-\frac{1}{x+1}\end{array}\right.$

Ta được:

$\int \frac{x\ln \left(x+1\right)}{{\left(x+1\right)}^2}\text{\hspace{0.33em}d}x=-\frac{x\ln \left(x+1\right)}{x+1}+\int \left[\frac{\ln \left(x+1\right)}{x+1}+\frac{x}{{\left(x+1\right)}^2}\right]\text{\hspace{0.33em}d}x$

$=-\frac{x\ln \left(x+1\right)}{x+1}+\int \left[\frac{\ln \left(x+1\right)}{x+1}+\frac{x+1-1}{{\left(x+1\right)}^2}\right]\text{\hspace{0.33em}d}x$

$=-\frac{x\ln \left(x+1\right)}{x+1}+\int \left[\frac{\ln \left(x+1\right)}{x+1}+\frac{1}{x+1}-\frac{1}{{\left(x+1\right)}^2}\right]\text{\hspace{0.33em}d}x$

$=-\frac{x\ln \left(x+1\right)}{x+1}+\frac{1}{2}{\ln }^2\left(x+1\right)+\ln \left(x+1\right)+\frac{1}{x+1}+C$