1708\hspace{-15}\mathbf{\int\frac{x^2-1}{(x^2+1).\sqrt{x^4+1}}dx=}$\\\\\\ $\mathbf{\int\frac{x^2-1}{x.(x+\frac{1}{x}).x.\sqrt{x^2+\frac{1}{x^2}}}dx}$\\\\\\ $\mathbf{\int\frac{(1-\frac{1}{x^2})}{(x+\frac{1}{x}).\sqrt{(x+\frac{1}{x})^2-(\sqrt{2})^2}}dx}$\\\\\\ Now Put $\mathbf{x+\frac{1}{x}=t\Leftrightarrow \left(1-\frac{1}{x^2}\right)dx=dt}$\\\\\\ $\mathbf{\int\frac{1}{t.\sqrt{t^2-(\sqrt{2})^2}}dt}$\\\\\\ $\mathbf{=\frac{1}{\sqrt{2}}.\boldsymbol{\sec}^{-1}\left(\frac{t}{\sqrt{2}}\right)+C}$\\\\\\ $\mathbf{=\frac{1}{\sqrt{2}}.\boldsymbol{\sec}^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)+C}$
1708
man111 singh
·2011-07-23 00:55:34
\hspace{-16}\mathbf{\int\sin^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx}$\\\\\\ $\mathbf{\int\sin^{-1}\left(\frac{2(x+1)}{\sqrt{4.(x+1)^2+9}}\right)dx}$\\\\\\ Now Put $\mathbf{2(x+1)=t\Leftrightarrow dx=\frac{1}{2}.dt}$\\\\\\ $\mathbf{=\frac{1}{2}.\int\sin^{-1}\left(\frac{t}{\sqrt{(t)^2+3^2}}\right)dt}$\\\\\\ Now Again Put $\mathbf{t=3.\tan\theta\Leftrightarrow dt=3\sec^2\theta d\theta}$ \\\\\\$\mathbf{=\frac{1}{2}.\int\sin^{-1}\left(\frac{3\tan\theta}{3.\sqrt{\tan^2\theta+1}}\right).3\sec^2\theta d\theta}$\\\\\\ $\mathbf{=\frac{3}{2}.\int\sin^{-1}(\sin\theta).\sec^2\theta d\theta}$\\\\\\ $\mathbf{=\frac{3}{2}.\int\theta.\sec^2\theta d\theta}$\\\\\\ $\mathbf{=\frac{3}{2}.\left(\theta.\tan\theta-\ln\left|\sec\theta\right|\right)+C}$\\\\\\ $\mathbf{=\frac{3}{2}.\left(\frac{t}{3}.\tan^{-1}\left(\frac{t}{3}\right)-\ln\left|\sqrt{1+\frac{t^2}{9}}\right|\right)+C}$\\\\\\
