JEE Main 2019 (Online) 11th January Evening ShiftQuestion 1
If $\int\frac{x+1}{\sqrt{2x-1}}dx=f(x)\sqrt{2x-1}+C$, where $C$ is a constant of integration, then $f(x)$ is equal to :
Indefinite Integration
Solution
Concept. Put $t=\sqrt{2x-1}$; the integrand becomes a polynomial in $t$. Formulas used. (1) $t=\sqrt{2x-1}\Rightarrow x=\dfrac{t^2+1}{2},\ dx=t\,dt$Step 1 - Substitute. By (1), $x+1=\dfrac{t^2+3}{2}$ and $$\int\frac{x+1}{t}\cdot t\,dt=\int\frac{t^2+3}{2}\,dt=\frac{t^3}{6}+\frac{3t}{2}+C.$$ Step 2 - Factor out $t$. $$t\left(\frac{t^2}{6}+\frac32\right)=t\cdot\frac{(2x-1)+9}{6}=\frac{x+4}{3}\sqrt{2x-1}.$$ So $f(x)=\frac13(x+4)$. Answer: (D)