INTEGRALS — Maths STD 12 Science — Question

Putting x + 2 = t

⇒ x = t - 2

$\frac{\text{dx}}{\text{dt}}=1$

⇒ dx = dt

Putting this value in eq. (i),

$\text{I}=\int\frac{3(\text{t}-2)-1}{(\text{t})^2}\text{dt}=\int\frac{3\text{t}-6-1}{\text{t}^2}\text{dt}=\int\frac{3\text{t}-7}{\text{t}^2}\text{dt}$

$=\int\Bigg(\frac{3\text{t}}{\text{t}^2}-\frac{7}{\text{t}^2}\Bigg)\text{dt}=\int\Bigg(\frac{3}{\text{t}}-\frac{7}{\text{t}^2}\Bigg)\text{dt}=3\int\frac{1}{\text{t}}\text{dt}-7\int\text{t}^{-2}\text{dt}$

$=3\text{log}|\text{t}|-7\frac{\text{t}^{-1}}{-1}+\text{c}=3\text{log}|\text{t}|+\frac{7}{\text{t}}+\text{c}=3\text{log}|\text{x}+2|+\frac{7}{\text{x}+2}+\text{c}$