If int f(x) dx = g(x) + c, then int f^{-1}(x) | vedclass

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

Question

(C) Let $y = f^{-1}(x)$, then $x = f(y)$.Differentiating both sides with respect to $y$, we get $dx = f'(y) dy$.Now, $\int f^{-1}(x) dx = \int y f'(y)

Vedclass · Accepted Answer

$x f^{-1}(x) - g(f^{-1}(x)) + c$

Vedclass · Answer

(C) Let $y = f^{-1}(x)$, then $x = f(y)$.Differentiating both sides with respect to $y$, we get $dx = f'(y) dy$.Now, $\int f^{-1}(x) dx = \int y f'(y) dy$.Using integration by parts: $\int u dv = uv - \int v du$, where $u = y$ and $dv = f'(y) dy$.Then $du = dy$ and $v = f(y)$.So, $\int y f'(y) dy = y f(y) - \int f(y) dy$.Since $\int f(y) dy = g(y) + c$, we have $y f(y) - g(y) + c$.Substituting $y = f^{-1}(x)$, we get $f^{-1}(x) \cdot f(f^{-1}(x)) - g(f^{-1}(x)) + c$.Since $f(f^{-1}(x)) = x$, the result is $x f^{-1}(x) - g(f^{-1}(x)) + c$.

If $\int f(x) dx = g(x) + c$, then $\int f^{-1}(x) dx =$

Similar Questions

$\int \frac{(1-4 \sin^2 x) \cos x}{\cos (3x+2)} dx =$

If $I_n = \int \tan^n x \, dx$ $(n > 1)$,then $I_4 + I_6 =$

$\int \frac{d x}{\left(x^2-a^2\right)^{\frac{3}{2}}}$ is equal to

$\int \frac{3\cos x + 3\sin x}{4\sin x + 5\cos x} \, dx = $

The value of $\frac{e^{-\pi/4} + \int_0^{\pi/4} e^{-x} \tan^{50} x \, dx}{\int_0^{\pi/4} e^{-x} (\tan^{49} x + \tan^{51} x) \, dx}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module