Evaluate the definite integral _ 0 ^ 1 2 x+3 5 x^ 2 +1 d x

Question

Evaluate the definite integral $\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$

✓

Answer

Let $I=\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1}$
Multiplying by $5$ in numerator and denominator
$I=\frac{1}{5} \int_{0}^{1} \frac{5(2 x+3)}{5 x^{2}+1} d x=\frac{1}{5} \int_{0}^{1} \frac{10 x+15}{5 x^{2}+1} d x$
$\Rightarrow I=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x+3 \int_{0}^{1} \frac{5}{5 x^{2}+1}$
$\Rightarrow I = I_1 + I_{2 }$
$I_{1}=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x$
Let $5x^2 + 1 = t …(i)$
$10x dx = dt …(ii)$
When $x = 0; t = 1$
When $x = 1; t = 6$
Substituting $(i)$ and $(ii)$ in $I_1$
$I_{1}=\frac{1}{5} \int_{1}^{6} \frac{d t}{t}=\frac{1}{5}[\log | t|]_{1}^{6}$
$I_{1}=\frac{1}{5}(\log |6|-\log |1|)=\frac{1}{5}(\log 6-0)$
$I_{1}=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x=\frac{\log 6}{5}$
$\mathrm{I}_{2}=3 \int {\frac{5}{5 \mathrm{x}^{2}+1}} \mathrm{dx}=\frac{3}{5} \int_{0}^{1} \frac{1}{\mathrm{x}^{2}+\frac{1}{5}} \mathrm{d} \mathrm{x} [\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c]$
$I_{2}=\frac{3}{5} \times \frac{1}{\frac{1}{\sqrt{5}}}\left[\tan ^{-1} \sqrt{5} x\right]_{0}^{1}=\frac{3}{5} \times \sqrt{5}\left(\tan ^{-1} \sqrt{5}-\tan ^{-1} 0\right)$
$I_2 = \frac{3}{\sqrt{5}}\tan ^{-1} \sqrt{5}$
$\because I = I_1 + I_2$
$\therefore \mathrm{I}=\frac{1}{5} \log 6+{\frac{3}{\sqrt{5}}} \tan ^{-1} \sqrt{5}$
$_{\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x = \frac{1}{5} \log 6+{\frac{3}{\sqrt{5}}} \tan ^{-1} \sqrt{5} }$