Mathematics

# $\int _{ 0 }^{ 1 }{ \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } } dx$

##### SOLUTION
$\displaystyle\int_{0}^{1}{\dfrac{xdx}{\sqrt{1+{x}^{2}}}}$
$=\dfrac{1}{2}\displaystyle\int_{0}^{1}{\dfrac{2xdx}{\sqrt{1+{x}^{2}}}}$
$=\dfrac{1}{2}\displaystyle\int_{0}^{1}{{\left(1+{x}^{2}\right)}^{\frac{-1}{2}}}$
$=\dfrac{1}{2}\left[\dfrac{{\left(1+{x}^{2}\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}\right]_{0}^{1}$
$=\dfrac{1}{2}\left[\dfrac{{\left(1+{x}^{2}\right)}^{\frac{3}{2}}}{\dfrac{3}{2}}\right]_{0}^{1}$
$=\dfrac{1}{2}\left[\dfrac{2{\left(1+{x}^{2}\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}$
$=\dfrac{1}{2}\left[\dfrac{2{\left(1+{1}^{2}\right)}^{\frac{3}{2}}}{3}-\dfrac{2{\left(1+0\right)}^{\frac{3}{2}}}{3}\right]$
$=\dfrac{1}{2}\left[\dfrac{{2}^{\frac{5}{2}}}{3}-\dfrac{2}{3}\right]$
$=\dfrac{1}{6}\left(2\left({2}^{\frac{3}{2}}-1\right)\right)$
$=\dfrac{{2}^{\frac{3}{2}}-1}{3}$
$=\dfrac{2\sqrt{2}-1}{3}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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