If I=∫1+x21−x2dx then I is equal to

I=∫1+x21−x2 1+x21+x2dx=∫1+x2dx1−x21+x2I=∫1dx1−x21+x2+∫x2dx1−x21+x2I=∫dx1−x21+x2−∫−x2dx1−x21+x2I=∫dx1−x21+x2−∫1−x2−1dx1−x21+x2I=∫11−x21+x2dx−∫1−x2dx1−x21+x2+∫1dx1−x21+x2I=2∫dx1−x21+x2−∫dx1+x2I=2∫dxx21x2−1x1+1x2−log⁡x+1+x2Put 1x2+1=t2 in 1st integral −2x3dx=dt2tdxx3=−tdt=I=2∫−tdtt2−2t−log⁡x+x2+1=I=2∫tdt(2)2−t2−log⁡x+x2+1+C=I=2⋅122log⁡2+t2−t−log⁡x+1+x2+C=I=12log⁡2+1x2+12−1x2+1−log⁡x+1+x2+C=I=12 log2x+1+x22x−1+x2 −logx+1+x2+C

If I=∫1+x21−x2dx then I is equal to

I=∫1+x21−x2 1+x21+x2dx=∫1+x2dx1−x21+x2I=∫1dx1−x21+x2+∫x2dx1−x21+x2I=∫dx1−x21+x2−∫−x2dx1−x21+x2I=∫dx1−x21+x2−∫1−x2−1dx1−x21+x2I=∫11−x21+x2dx−∫1−x2dx1−x21+x2+∫1dx1−x21+x2I=2∫dx1−x21+x2−∫dx1+x2I=2∫dxx21x2−1x1+1x2−log⁡x+1+x2Put 1x2+1=t2 in 1st integral −2x3dx=dt2tdxx3=−tdt=I=2∫−tdtt2−2t−log⁡x+x2+1=I=2∫tdt(2)2−t2−log⁡x+x2+1+C=I=2⋅122log⁡2+t2−t−log⁡x+1+x2+C=I=12log⁡2+1x2+12−1x2+1−log⁡x+1+x2+C=I=12 log2x+1+x22x−1+x2 −logx+1+x2+C

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$If I = \int \frac{\sqrt{1 + x^{2}}}{1 - x^{2}} d x then I is equal to$

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$\frac{- 1}{\sqrt{2}} \log |\frac{\sqrt{2} x + \sqrt{1 + x^{2}}}{\sqrt{2} x - \sqrt{1 + x^{2}}}| - \log |x + \sqrt{1 + x^{2}}| + C$

$\frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} x + \sqrt{1 + x^{2}}}{\sqrt{2} x - \sqrt{1 + x^{2}}}| - \log |x + \sqrt{1 + x^{2}}| + C$

$\sqrt{2} \log |\frac{\sqrt{2} x + \sqrt{1 + x^{2}}}{\sqrt{2} x - \sqrt{1 + x^{2}}}| - \log |x + \sqrt{1 + x^{2}}| + C$

$\frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} x + \sqrt{1 + x^{2}}}{\sqrt{2} x - \sqrt{1 + x^{2}}}| + \log |x + \sqrt{1 + x^{2}}| + C$

answer is D.

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Detailed Solution

$I = \int \frac{\sqrt{1 + x^{2}}}{1 - x^{2}} \sqrt{\frac{1 + x^{2}}{1 + x^{2}}} d x = \int \frac{(1 + x^{2}) d x}{(1 - x^{2}) \sqrt{1 + x^{2}}}$

$\begin{aligned} I = \int \frac{1 d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} + \int \frac{x^{2} d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} \\ I = \int \frac{d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} - \int \frac{- x^{2} d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} \end{aligned}$

$\begin{array}{l} I = \int \frac{d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} - \int \frac{[(1 - x^{2}) - 1] d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} \\ I = \int \frac{1}{(1 - x^{2}) \sqrt{1 + x^{2}}} d x - \int \frac{(1 - x^{2}) d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} + \int \frac{1 d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} \\ I = 2 \int \frac{d x}{(1 - x^{2}) \sqrt{1 + x^{2}}} - \int \frac{d x}{\sqrt{1 + x^{2}}} \\ I = 2 \int \frac{d x}{x^{2} (\frac{1}{x^{2}} - 1) x \sqrt{1 + \frac{1}{x^{2}}}} - \log |x + \sqrt{1 + x^{2}}| \end{array}$

$Put \frac{1}{x^{2}} + 1 = t^{2} in 1^{st} integral$

$\begin{array}{l} - \frac{2}{x^{3}} d x = d t 2 t \\ \frac{d x}{x^{3}} = - t d t \\ = I = 2 \int \frac{- t d t}{(t^{2} - 2) t} - \log |x + \sqrt{x^{2} + 1}| \\ = I = 2 \int \frac{t d t}{(\sqrt{2})^{2} - t^{2}} - \log |x + \sqrt{x^{2} + 1}| + C \\ = I = 2 \cdot \frac{1}{2 \sqrt{2}} \log |\frac{\sqrt{2} + t}{\sqrt{2} - t}| - \log |x + \sqrt{1 + x^{2}}| + C \\ = I = \frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} + \sqrt{\frac{1}{x^{2}} + 1}}{\sqrt{2} - \sqrt{\frac{1}{x^{2}} + 1}}| - \log |x + \sqrt{1 + x^{2}}| + C \end{array}$