MHT CET · Maths · Indefinite Integration
\(\int \mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}+\mathrm{x}\right)} \mathrm{dx}=\)
- A \(\mathrm{e}^{\mathrm{x}}+\mathrm{x}+\mathrm{c}\)
- B \(\mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}\right)} \cdot \mathrm{x}+\mathrm{cd}\)
- C \(\mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}\right)}+\mathrm{c}\)
- D \(\mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}\right)}\left(\mathrm{e}^{\mathrm{x}}-1\right)+\mathrm{c}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}\right)}+\mathrm{c}\)
Step-by-step Solution
Detailed explanation
\(
I=\int e^{e^x+x} d x=\int e^{e^x} \cdot e^x d x
\)
Put \(\mathrm{e}^{\mathrm{x}}=\mathrm{t} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\)
\(
\therefore \mathrm{I}=\int \mathrm{e}^{\mathrm{t}} \mathrm{dt}=\mathrm{e}^{\mathrm{t}}+\mathrm{c}=\mathrm{e}^{\mathrm{e}^{\mathrm{x}}}+\mathrm{c}
\)
I=\int e^{e^x+x} d x=\int e^{e^x} \cdot e^x d x
\)
Put \(\mathrm{e}^{\mathrm{x}}=\mathrm{t} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\)
\(
\therefore \mathrm{I}=\int \mathrm{e}^{\mathrm{t}} \mathrm{dt}=\mathrm{e}^{\mathrm{t}}+\mathrm{c}=\mathrm{e}^{\mathrm{e}^{\mathrm{x}}}+\mathrm{c}
\)
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