If {f}(x)=1+x ; {g}(x)= x, then {g}({f}(x)) d x is equal to

Correct Answer is : (1+x)log⁡(1+x)−x+c,(1+x) (1+x)-x+c_{,}(1+x)log(1+x)−x+c,(where c is a constant of integration)

(1+x) (1+x)-x+c_{,}(where c is a constant of integration)

(1+x) x-x+c_{,}(where c is a constant of integration)

x (1+x)+c_{,}(where c is a constant of integration)

(1+x) (1+x)+x+c, (where c is a constant of integration)

Correct Answer is : (1+x)log⁡(1+x)−x+c,(1+x) (1+x)-x+c_{,}(1+x)log(1+x)−x+c,(where c is a constant of integration)

Test Index

MHTCET 11 May 2024 Shift 2 Paper

Question : 124 of 150

Marks: +1, -0

\mathrm{f}(x)=1+x ; \mathrm{g}(x)=\log x

\int \mathrm{g}(\mathrm{f}(x)) d x

is equal to

Solution:

Concept:

Use the formula for integration by parts:

\int u \, dv = uv - \int v \, du

Explanation:

Given

\mathrm{f}(x)=1+x

and

\mathrm{g}(x)=\log x

\mathrm{g}(\mathrm{f}(x)) = \log(1+x)

Thus we need to evaluate

\int \log(1+x) \, dx

Write it as

\int 1 \cdot \log(1+x) \, dx

and take

u=\log(1+x)

dv=dx

Then

du = \frac{1}{1+x} dx

and

v = x

Applying integration by parts:

\int \log(1+x) \, dx = x \log(1+x) - \int \frac{x}{1+x} \, dx + c.

Now rewrite

\frac{x}{1+x} = 1 - \frac{1}{1+x}

So the integral becomes:

x \log(1+x) - \left[ \int 1 \, dx - \int \frac{1}{1+x} \, dx \right] + c.

This simplifies to:

x \log(1+x) - x + \log(1+x) + c.

Factor out

\log(1+x)

(1+x)\log(1+x) - x + c.

Answer:

\int \mathrm{g}(\mathrm{f}(x)) \, dx = (1+x)\log(1+x) - x + c.

Hence the correct option is A.

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