Indefinite Integrals — Maths STD 12 Science — Question

let $2\text{x}+5=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x}+5)+\mu$

$=\mu(2\text{x}+2)+\mu$

$2\text{x}+5=(2\lambda)\text{x}+2\lambda+\mu$

Compairing the coefficient of like powewrs of x,

$2\lambda=2\ \Rightarrow\lambda=1$

$2\lambda+\mu=5\ \Rightarrow 2(1)+\mu=5$

$\Rightarrow\mu=3$

So, $\text{I}=\int\frac{(2\text{x}+2)+3}{\sqrt{\text{x}^2+2\text{x}+5}}\text{dx}$

$=\int\frac{(2\text{x}+3)}{\sqrt{\text{x}^2+2\text{x}+5}}\text{dx}+3\int\frac{1}{\sqrt{\text{x}^2+2\text{x}+(1)^2-(1)^2+5}}\text{dx}$

$\text{I}=\int\frac{2\text{x}+3}{\sqrt{\text{x}^2+2\text{x}+5}}+3\int\frac{1}{\sqrt{(\text{x}+1)^2+(2)^2}}\text{dx}$

$\text{I}=2\sqrt{\text{x}^22\text{x}+5}+3\log\big|\text{x}+1+{\sqrt{(\text{x}+1)^2+(2)^2}}\big|+\text{C}$ $\big[\text{since}, \int\frac{1}{\text{x}}\text{dx}=2\sqrt{\text{x}+\text{C},\int\frac{1}{\sqrt{\text{x}^2+\text{a}^2}}}\text{dx}=\log\big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}+\text{C}\big]$

$\text{I}=2\sqrt{\text{x}^2+2\text{x}+5}+3\log\big|\text{x}+1+\sqrt{\text{x}^2+2\text{x}+5}\big|+\text{C}$