Question

Solve ${\int }_0^1\frac{2x+3}{5x^2+1}dx$

• A. $\frac{1}{5}log6-\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}$
• B. $\frac{1}{5}log6+\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}$
• C. $log6-\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}$
• D. $log6+\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}$

Answer 1

Answer: (B)

$\frac{1}{5}log6+\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}$

$I={\int }_0^1\frac{2x+3}{5x^2+1}dx$

$I={\int }_0^1\frac{2x}{5x^2+1}dx+{\int }_0^1\frac{3}{5x^2+1}dx$

$I_1={\int }_0^1\frac{2x}{5x^2+1}dx$ and $I_2={\int }_0^1\frac{3}{5x^2+1}dx$

Now, Let $5x^2+1=t$

$10xdx=dt$

$xdx=\frac{1}{10}dt$

$\therefore I_1={\int }_0^1\frac{2x}{5x^2+1}={\int }_1^6\frac{2}{t}\frac{dt}{10}$

$I_1=\frac{1}{5}{\int }_1^6\frac{dt}{t}$

$I_1=\frac{1}{5}\log t{|}_1^6$

$I_1=\frac{1}{5}\log 6$

Now,$I_2={\int }_0^1\frac{3}{5x^2+1}dx=\frac{3}{5}{\int }_0^1\frac{1}{x^2+\frac{1}{5}}dx$

$I_2=\frac{3}{5}\frac{1}{\frac{1}{\sqrt{5}}}{\tan }^{-1}\left(\frac{x}{\frac{1}{\sqrt{5}}}\right)=\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}x{|}_0^1$

$I_2=\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}\left(1\right)-\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}\left(0\right)=\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}$

$I=I_1+I_2=\frac{1}{5}\log 6+\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}$