Mathematics

Evaluate : $$\displaystyle \int_{0}^{1}\dfrac {2x + 3}{5x^{2} + 1} dx$$


SOLUTION
$$I=\displaystyle \int_{0}^{1}\dfrac{2x+3}{5x^2+1}dx=\int_{0}^{1}\dfrac{2x}{5x^2+1}dx+\int_{0}^{1}\dfrac{3}{5x^2+1}dx$$
$$\Rightarrow I=I_1+I_2$$
$$I_1=\displaystyle \int_{0}^{1}\dfrac{2x}{5x^2+1}dx$$
Put $$5x^2+1=t\Rightarrow 5\times 2xdx=dt$$ and limits will be 
Lower limit $$x=0\Rightarrow t=1$$ and Upper limit $$x=1\Rightarrow t=6$$
$$I_1=\displaystyle \dfrac{1}{5}\int_{1}^{6}\dfrac{1}{t}dt=\dfrac{1}{5}|\ln|t||_1^{6}=\dfrac{1}{5}|\ln 6-\ln 1|=\dfrac{1}{5}\ln 6$$
$$I_2=\displaystyle \int_{0}^{1}\dfrac{3}{5x^2+1}dx=\dfrac{3}{5}\int_{0}^{1}\dfrac{1}{x^2+\dfrac{1}{5}}dx$$
Integral of $$\displaystyle \int_{a}^{b}\dfrac{1}{x^2+a^2}dx=\left | \dfrac{1}{a}\tan^{-1}\dfrac{x}{a}\right |_{a}^{b}$$
Here, $$a=\dfrac{1}{\sqrt{5}}$$
$$I_2=\dfrac{3}{5}\left |\sqrt{5}\tan^{-1}(\sqrt{5}x) \right |_{0}^{1}=\dfrac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5}) $$
$$\therefore I= \dfrac{1}{5}\ln 6+\dfrac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$$
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