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Let J=\int \frac{\ln x}{x^{10}} d x.\therefore By product Rule,\begin{aligned}\int f g^{\prime} & =f g-\int f^{\prime} g \\f & =\ln x \quad g^{\prime}=\frac{1}{x^{10}} \\f^{\prime} & =\frac{1}{x} \\\therefore \frac{\ln x}{x^{10}} d x & =\frac{-\ln x}{9 x^{9}}-\frac{-1}{9 x^{9}}-\frac{1}{9 x^{10}} d x \\& =\frac{-\ln x}{9 x^{9}}-\frac{1}{81 x^{9}}+c \\& =-\frac{9 \ln x+1}{81 x^{9}}+c\end{aligned}112) Let J=\int x^{5} \cos \left(x^{2}\right) d x.put\begin{array}{l}u=x^{2} \\d u=2 x d x\end{array}d x=\frac{1}{2 x} d u .x^{4}=4^{2}\begin{array}{l}\therefore \int x^{5} \cos \left(x^{2}\right) d x=\frac{1}{2} \int u^{2} \cos u d u . \\\therefore I=\frac{1}{2} \int u^{2} \cos u d u .\end{array}\therefore product Rule.\begin{array}{ll}\int f g^{\prime}=f g-\int f_{g} g \\f=u^{2} & g^{\prime}=\cos u \\f^{\prime}=2 u & g=\sin u .\end{array}\begin{aligned}\frac{1}{2} \int u^{2} \cos u d u & =\frac{1}{2}\left[u^{2} \sin u-\int 2 u \sin u d u\right] . \\& =\frac{1}{2}\left[u^{2} \sin u-\left[-u \cos u+\int \cos u d u\right] \cdots[\text { Again by pundudt] pute }\right.\end{aligned}\begin{array}{l}=\frac{1}{2}\left[u^{2} \sin u+2 u \cos u-2 \sin u\right]+c . \\=\frac{u^{2} \sin u}{2}-\sin u+u \cos u+c\end{array}put value \quad x^{2}=uI=\frac{x^{4} \sin \left(x^{2}\right)}{2}-\sin \left(x^{2}\right)+x^{2} \cos \left(x^{2}\right)+c \text {. } ...