2 câu trả lời
Đáp án:
$\frac{x^2}{2}$ .ln(1+x)-$\frac{x^2}{4}$+$\frac{x}{2}$ -$\frac{1}{2}$ln(1+x) +C.
Giải thích các bước giải:
du=$\frac{1}{1+x}$ dx, v=$\frac{x^2}{2}$ →∫xln(1+x)dx=uv-∫vdu
=$\frac{x^2}{2}$ .ln(1+x)-∫$\frac{x^2}{2(1+x)}$ dx
=$\frac{x^2}{2}$ .ln(1+x)-$\frac{1}{2}$ ∫$\frac{x^2}{1+x}$ dx
=$\frac{x^2}{2}$ .ln(1+x)-$\frac{1}{2}$ ∫$\frac{(x+1)(x-1)}{1+x}$dx -$\frac{1}{2}$ ∫$\frac{1}{1+x}$ dx
=$\frac{x^2}{2}$ .ln(1+x)-$\frac{1}{2}$ ∫(x-1)dx-$\frac{1}{2}$ln(1+x)
=$\frac{x^2}{2}$ .ln(1+x)-$\frac{1}{2}$ .($\frac{x^2}{2}$ -x)-$\frac{1}{2}$ln(1+x) +C
=$\frac{x^2}{2}$ .ln(1+x)-$\frac{x^2}{4}$+$\frac{x}{2}$ -$\frac{1}{2}$ln(1+x) +C.
Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
u = \ln \left( {1 + x} \right) \Rightarrow du = \frac{1}{{1 + x}}dx\\
dv = xdx \Rightarrow v = \frac{{{x^2}}}{2}
\end{array} \right.\\
\Rightarrow \int {x\ln \left( {1 + x} \right)dx} \\
= u.v - \int {v.du} \\
= \ln \left( {x + 1} \right).\frac{{{x^2}}}{2} - \int {\frac{{{x^2}}}{2}.\frac{1}{{x + 1}}dx} \\
= \ln \left( {x + 1} \right).\frac{{{x^2}}}{2} - \frac{1}{2}\int {\frac{{{x^2} - 1 + 1}}{{x + 1}}dx} \\
= \ln \left( {x + 1} \right).\frac{{{x^2}}}{2} - \frac{1}{2}.\int {x - 1 + \frac{1}{{x + 1}}dx} \\
= \ln \left( {x + 1} \right).\frac{{{x^2}}}{2} - \frac{1}{2}.\left( {\frac{{{x^2}}}{2} - x + \ln \left( {x + 1} \right)} \right)\\
= \frac{1}{2}\ln \left( {x + 1} \right).\left( {{x^2} - 1} \right) - \frac{{{x^2}}}{4} + \frac{x}{2} + C
\end{array}$
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