Tích phân từng phần

$1,  \int\limits_{0}^{\frac{\pi}{2}}(x+\cos^3x)\sin xdx$

2, $\int\limits_{0}^{2}(2x+7)\ln(x+1)dx$

3,  $\int\limits_{2}^{3}\ln(x^2-x)dx$
1. Ta có:
      $\int\limits_0^{\frac{\pi}{2}}(x+\cos^3x)\sin xdx$
$=\int\limits_0^{\frac{\pi}{2}}\cos^3x\sin xdx+\int\limits_0^{\frac{\pi}{2}}x\sin xdx$
$=-\int\limits_0^{\frac{\pi}{2}}\cos^3xd(\cos x)-\int\limits_0^{\frac{\pi}{2}}xd(\cos x)$
$=-\dfrac{\cos^4x}{4}\left|\begin{array}{l}\dfrac{\pi}{2}\\0\end{array}\right.-x\cos x\left|\begin{array}{l}\dfrac{\pi}{2}\\0\end{array}\right.+\int\limits_0^{\frac{\pi}{2}}\cos xdx$
$=\dfrac{1}{4}+\sin x \left|\begin{array}{l}\dfrac{\pi}{2}\\0\end{array}\right.$
$=\dfrac{5}{4}$
Bài 3. Đặt $\ln (x^2-x) =u \Rightarrow \dfrac{2x-1}{x^2-x}dx=du$  và $dx=dv \Rightarrow x=v$

$I=x\ln (x^2-x) -\int x \dfrac{2x-1}{x^2-x}dx = x\ln (x^2-x) \int \dfrac{2x-1}{x-1}dx$

$=x\ln (x^2 -x) -\int \bigg ( 2+\dfrac{1}{x-1} \bigg ) dx$ đơn giản rồi nhé
Bài 2 đặt $\ln (x+1)= u \Rightarrow \dfrac{1}{x+1}dx = du$  và  $(2x+7)dx = dv \Rightarrow x^2 +7x = v$

$I= (x^2+7x)\ln (x+1) - \int \dfrac{x^2+7x}{x+1}dx= (x^2+7x)\ln (x+1) - \int \bigg ( x+ 6 -\dfrac{6}{x+1} \bigg ) dx$

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