If int frac{x+1}{sqrt{2x-1}} , dx = f(x) sqrt | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int \frac{x+1}{\sqrt{2x-1}} \, dx$. Substitute $2x-1 = t^2$,which implies $x = \frac{t^2+1}{2}$ and $dx = t \, dt$. Then $x+1 = \frac{t^

Vedclass · Accepted Answer

$\frac{1}{3}(x+4)$

Vedclass · Answer

(B) Let $I = \int \frac{x+1}{\sqrt{2x-1}} \, dx$. Substitute $2x-1 = t^2$,which implies $x = \frac{t^2+1}{2}$ and $dx = t \, dt$. Then $x+1 = \frac{t^2+1}{2} + 1 = \frac{t^2+3}{2}$. Substituting these into the integral: $I = \int \frac{(\frac{t^2+3}{2}) t \, dt}{t} = \int \frac{t^2+3}{2} \, dt$. $I = \frac{1}{2} (\frac{t^3}{3} + 3t) + c = \frac{t^3}{6} + \frac{3t}{2} + c$. Factor out $\frac{t}{6}$: $I = \frac{t}{6} (t^2 + 9) + c$. Substitute $t = \sqrt{2x-1}$ back: $I = \frac{\sqrt{2x-1}}{6} (2x-1+9) + c = \frac{\sqrt{2x-1}}{6} (2x+8) + c$. $I = \sqrt{2x-1} \cdot \frac{2(x+4)}{6} + c = \sqrt{2x-1} \cdot \frac{x+4}{3} + c$. Comparing this with $f(x) \sqrt{2x-1} + c$,we get $f(x) = \frac{x+4}{3}$.

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

Similar Questions

Find the integral of the function $\sin ^{3} x \cos ^{3} x$.

$\int \frac{2 \tan (x)}{1+2 \tan ^2(x)} d x=$

$\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{1+x^{100}} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+c$, then $k$ is equal to

Consider $f(x) = \frac{x^2}{1 + x^3}$ and $g(t) = \int f(t) \, dt$. If $g(1) = 0$,then $g(x)$ equals:

Integrate the function: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module