tính tích phâna, $I=\int\limits_{1}^{2} \frac{\ln( x+1)}{x^{2}}dx $
                        b,$I=\int\limits_{0}^{\frac{\pi}{2}}(x+1)\sin x.\cos 2xdx $   
                         c,$I=\int\limits_{2}^{3} \ln (x^{2}-4x+5)dx$
a,
Ta có:
$I=\int\limits_{1}^{2} \frac{\ln( x+1)}{x^{2}}dx$
    $=\int\limits_{2}^{1}\ln( x+1)d(\frac{1}{x})$
    $=\frac{\ln(x+1)}{x} \left|\begin{array}{l}1\\2\end{array}\right.-\int\limits_{2}^{1} \frac{d(\ln( x+1))}{x}$
    $=\ln2-\frac{1}{2}\ln3-\int\limits_2^1\frac{dx}{x(x+1)}$
    $=\ln2-\frac{1}{2}\ln3-\int\limits_2^1\left(\frac{1}{x}-\frac{1}{x+1}\right)dx$
    $=\ln2-\frac{1}{2}\ln3-\ln\frac{x}{x+1} \left|\begin{array}{l}1\\2\end{array}\right. $
    $=2\ln2-\frac{1}{2}\ln3+\ln\frac{2}{3}=\ln\frac{8}{3\sqrt3}$
Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! –  khangnguyenthanh 16-01-13 06:06 PM
em cảm ơn rất nhiều –  linhdajka95 15-01-13 08:22 PM
b,
Ta có:
$I=\int\limits_{0}^{\frac{\pi}{2}}(x+1)\sin x\cos2xdx$
    $=\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}(x+1)\sin3xdx-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}(x+1)\sin xdx$
    $=-\frac{1}{6}\int\limits_{0}^{\frac{\pi}{2}}(x+1)d(\cos3x)+\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}(x+1)d(\cos x)$
    $=-\frac{1}{6}(x+1)\cos3x \left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.+\frac{1}{6}\int\limits_{0}^{\frac{\pi}{2}}\cos 3xd(x+1)+\frac{1}{2}(x+1)\cos x \left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\cos xd(x+1)$
    $=\frac{1}{6}-\frac{1}{2}+\frac{1}{6}\int\limits_{0}^{\frac{\pi}{2}}\cos 3xdx-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\cos xdx$
    $=-\frac{1}{3}+\frac{\sin3x}{18} \left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.-\frac{\sin x}{2} \left|\begin{array}{l}\frac{\pi}{2}\\0\end{array}\right.$
    $=-\frac{1}{3}-\frac{1}{18}-\frac{1}{2}=-\frac{8}{9}$
Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! –  khangnguyenthanh 16-01-13 06:06 PM
e rất cảm ơn a –  linhdajka95 15-01-13 08:23 PM

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