Question

Evaluate the definite integrals.
${\int }_0^1\frac{2x+3}{(5x^2+1)}dx.$

Answer 1

Let $I={\int }_0^1\frac{2x+3}{(5x^2+1)}dx={\int }_0^1\frac{2x}{(5x^2+1)}dx+3{\int }_0^1\frac{1}{(5x^2+1)}dx$
Put $5x^2+1=t\Rightarrow 10x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{10x}$
$\therefore I={\int }_0^1\frac{2x}{t}\frac{dt}{10x}+\frac{3}{5}{\int }_0^1\frac{1}{x^2+(\frac{1}{\sqrt{5}}{)}^2}dx=\frac{1}{5}{\int }_0^1\frac{1}{t}dt+\frac{3}{5}{\int }_0^1\frac{1}{x^2+(\frac{1}{\sqrt{5}}{)}^2}dx=\frac{1}{5}[logt{]}_0^1+\frac{3}{5}\times \frac{1}{\frac{1}{\sqrt{5}}}{\left[ta{n}^{-1}\frac{x}{\frac{1}{\sqrt{5}}}\right]}_0^1(\because \int \frac{dx}{a^2+x^2}=\frac{1}{a}ta{n}^{-1}\frac{x}{a})=\frac{1}{5}[log|5x^2+1|{]}_0^1+\frac{3}{5}\sqrt{5}[ta{n}^{-1}\sqrt{5}x{]}_0^1=\frac{1}{5}[log|5+1|-log|0+1|]+\frac{3}{\sqrt{5}}[ta{n}^{-1}\sqrt{5}\left(1\right)-ta{n}^{-1}\sqrt{5}\left(0\right)]=\frac{1}{5}log6+\frac{3}{\sqrt{5}}ta{n}^{-1}\sqrt{5}(\because log1=0andta{n}^{-1}\left(0\right)=0)$