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
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