MHT CET · Maths · Indefinite Integration
If \(\mathrm{f}(x)=1+x ; \mathrm{g}(x)=\log x\), then \(\int \mathrm{g}(\mathrm{f}(x)) \mathrm{d} x\) is equal to
- A \((1+x) \log (1+x)-x+\mathrm{c}\), (where c is a constant of integration)
- B \((1+x) \log x-x+\mathrm{c}\), (where c is a constant of integration)
- C \(x \log (1+x)+\mathrm{c}\), (where c is a constant of integration)
- D \((1+x) \log (1+x)+x+\mathrm{c}\), (where c is a constant of integration)
Answer & Solution
Correct Answer
(A) \((1+x) \log (1+x)-x+\mathrm{c}\), (where c is a constant of integration)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int \mathrm{g}(\mathrm{f}(x)) \mathrm{d} x \\ & =\int 1 \times \log (1+x) \mathrm{d} x\end{aligned}\)
\(\begin{aligned} & =x \log (1+x)-\int x \times \frac{1}{(1+x)} \mathrm{d} x+\mathrm{c} \\ & =x \log (1+x)-\left[\int \frac{1+x}{1+x} \mathrm{~d} x-\int \frac{1}{1+x} \mathrm{~d} x\right]+\mathrm{c} \\ & =x \log (1+x)-x+\log (1+x)+\mathrm{c} \\ & =(1+x) \log (1+x)-x+\mathrm{c}\end{aligned}\)
\(\begin{aligned} & =x \log (1+x)-\int x \times \frac{1}{(1+x)} \mathrm{d} x+\mathrm{c} \\ & =x \log (1+x)-\left[\int \frac{1+x}{1+x} \mathrm{~d} x-\int \frac{1}{1+x} \mathrm{~d} x\right]+\mathrm{c} \\ & =x \log (1+x)-x+\log (1+x)+\mathrm{c} \\ & =(1+x) \log (1+x)-x+\mathrm{c}\end{aligned}\)
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