Indefinite Integrals — Maths STD 12 Science — Question

Let $2\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x}-1)+\mu$

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

$2\text{x}+1=(2\lambda)\text{x}+2\lambda+\mu$

Comapring the coefficient of like powers of x,

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

$2\lambda+\mu=1\Rightarrow2(1)+\mu=1$

$\mu=-1$

So, $\text{I}=\int\frac{(2\text{x}+2)-1}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}$

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

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

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

$\text{I}=2\sqrt{\text{x}^2+2\text{x}-1}-\log\big|\text{x}+1+\sqrt{\text{x}^2+2\text{x}-1}\big|+\text{c}$