12) ∫1xn+xdx =∫1xn(1+x1-n)dx ; now let 1+x1-n =z or,(1-n)1xndx=dz thus, 1(1-n)∫1zdz=1(1-n)lnz +c =1(1-n)ln(1+x1-n) +c.
\hspace{-16}(1)\;\;\mathbf{\int_{0}^{\pi}\frac{x^2\cos^2 x-x\sin x-\cos x-1}{(1+x\sin x)^2}dx}\\\\\\ (2)\;\; \mathbf{\int\frac{1}{x^n+x}dx}\\\\\\ (3)\;\;\mathbf{\int\frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx}
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113)∫11+√x+√(x+1)dx=∫1+√x-√(x+1)1+2√x+x-x-1dx=∫1+√x-√(x+1)2√xdx=√x+12x+12∫√(x+1x)dx.now let √x=z or,1√xdx=2dz. and √(x+1)=√(z2+1).thus, the integral=√x+12x+∫√(z2+1)dz. now it can be done easily.
1708Thanks aritra.
\hspace{-16}(1)\;\;\displaystyle\int_{0}^{\pi}\frac{x^2\cos^2 x-x\sin x-\cos x-1}{(1+x\sin x)^2}dx$\\\\\\ $=\int_{0}^{\pi}\frac{x^2-x^2\sin^2 x-x\sin x-\cos x -1}{(1+x\sin x)^2}dx$\\\\\\ $=\int_{0}^{\pi}\frac{(-x^2\sin^2 x-1-2x\sin x)-(1-x\sin x-x^2)}{(1+x\sin x)^2}dx$\\\\\\ $=\int_{0}^{\pi}-1.dx-\int_{0}^{\pi}\frac{d}{dx}\left(\frac{x.\cos x}{1+\sin x}\right).dx$\\\\\\ $=-\left[x\right]_{0}^{\pi}-\left[\frac{x.\cos x}{(1+x.\sin x)}\right]_{0}^{\pi}$\\\\\\ $=-\pi+\pi=0$