Bài 3.14 trang 166 SBT giải tích 12

Đề bài

$\int {x\ln \left( {x + 1} \right)dx}$ bằng

A. $\left( {\dfrac{{{x^2}}}{2} - 1} \right)\ln \left( {x + 1} \right) + \dfrac{1}{4}{\left( {x - 1} \right)^2} + C$

B. $\left( {\dfrac{{{x^2}}}{2} - 1} \right)\ln \left( {x + 1} \right) - \dfrac{1}{2}{\left( {x - 1} \right)^2} + C$

C. $\left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left( {x + 1} \right) - \dfrac{1}{4}{\left( {x - 1} \right)^2} + C$

D. $\left( {\dfrac{{{x^2}}}{2} + 1} \right)\ln \left( {x + 1} \right) - \dfrac{1}{4}{\left( {x - 1} \right)^2} + C$

Lời giải chi tiết

Đặt $\left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\dv = xdx\end{array} \right.$ $\Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{{x + 1}}dx\\v = \dfrac{{{x^2}}}{2}\end{array} \right.$

Khi đó $\int {x\ln \left( {x + 1} \right)dx}$$= \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{1}{2}\int {\dfrac{{{x^2}}}{{x + 1}}dx}$ $= \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{1}{2}\int {\left( {x - 1 + \dfrac{1}{{x + 1}}} \right)dx}$

$= \dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right) - \dfrac{{{x^2}}}{4} + \dfrac{1}{2}x - \dfrac{1}{2}\ln \left( {x + 1} \right) + C$ $= \left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left( {x + 1} \right) - \dfrac{1}{4}\left( {{x^2} - 2x + 1} \right) + C'$

$= \left( {\dfrac{{{x^2}}}{2} - \dfrac{1}{2}} \right)\ln \left( {x + 1} \right) - \dfrac{1}{4}{\left( {x - 1} \right)^2} + C'$

Chọn C.

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