If f(x)={ ^{-1} x}{{1-x²}} and g(x)=e^{ ^{-1} x}, then the value of f(x) g(x) d x=

Correct Answer is : esin⁡−1x(sin⁡−1x−1)+ce^{ ^{-1} x}≤ft( ^{-1} x-1)+cesin−1x(sin−1x−1)+c

e^{ ^{-1} x}≤ft( ^{-1} x-1)+c

e^{ ^{-1} x}≤ft(1- ^{-1} x)+c

e^{ ^{-1} x}≤ft( ^{-1} x+1)+c

e^{ ^{-1} x}≤ft(- ^{-1} x-1)+c

Correct Answer is : esin⁡−1x(sin⁡−1x−1)+ce^{ ^{-1} x}≤ft( ^{-1} x-1)+cesin−1x(sin−1x−1)+c

Test Index

MHTCET 17 Apr 2026 Shift 2 Paper

Question : 130 of 150

Marks: +1, -0

f(x)=\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}

g(x)=e^{\sin ^{-1} x}

\int f(x) g(x) d x=\ldots

Solution:

Concept:

Use substitution to simplify the integrand, then apply integration by parts.

Explanation:

We have

f(x)g(x)=\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}e^{\sin^{-1}x}

Let

t=\sin^{-1}x \quad \Rightarrow \quad dt=\frac{dx}{\sqrt{1-x^{2}}}

Therefore,

\int f(x)g(x)\,dx=\int te^{t}\,dt

Using integration by parts,

\int te^{t}\,dt=te^{t}-e^{t}+C=e^{t}(t-1)+C

Substituting back

t=\sin^{-1}x

\int f(x)g(x)\,dx=e^{\sin^{-1}x}\left(\sin^{-1}x-1\right)+C

Answer:

Option A:

e^{\sin ^{-1} x}\left(\sin ^{-1} x-1\right)+c

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