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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2017 · Mathematics · Indefinite Integration

What is ∫ ( ^ e-1 + ^ x-1 ) dx ^ e + ^ x equal to?

What is ∫(\mathrmxe−1+\mathrmex−1)dx\mathrmxe+\mathrmex\int \frac{\left(\mathrmx^{\mathrm{e}-1}+\mathrme^{\mathrm{x}-1}\right) \mathrm{dx}}{\mathrmx^{\mathrm{e}}+\mathrme^{\mathrm{x}}} equal to?

Options

  1. Ax22+c\frac{\mathrm{x}^2}{2}+\mathrm{c}
  2. BIn⁡(x+e)+c\operatorname{In}(x+e)+c
  3. C\operatorname{In}(\mathrmxe+\mathrmex)+c\left(\mathrmx^{\mathrm{e}}+\mathrme^{\mathrm{x}}\right)+\mathrm{c}
  4. D\frac{1}{\mathrm{e}} \operatorname{In}(\mathrmxe+\mathrmex)+c\left(\mathrmx^{\mathrm{e}}+\mathrme^{\mathrm{x}}\right)+\mathrm{c}

Correct answer

D. \frac{1}{\mathrm{e}} \operatorname{In}(\mathrmxe+\mathrmex)+c\left(\mathrmx^{\mathrm{e}}+\mathrme^{\mathrm{x}}\right)+\mathrm{c}

Step-by-step solution

∫(xe−1+ex−1)dxxe+exPut xe+ex=t⇒exee−1+ex=dtdx⇒(exee−1+ex)dx=dt∴∫(xe−1+ex−1)dxxe+ex=1e∫e⋅xe−1+exxe+ex⋅dx=1e∫dtt=1e⋅ln⁡t+c=1e⋅ln⁡(xe+ex)+c\begin{aligned} & \int \frac{\left(\mathrm{x}^{\mathrm{e}-1}+\mathrm{e}^{\mathrm{x}-1}\right) \mathrm{dx}}{\mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}} \\ & \text{Put } \mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}=\mathrm{t} \Rightarrow \mathrm{ex} \mathrm{e}^{\mathrm{e}-1}+\mathrm{e}^{\mathrm{x}}=\frac{\mathrm{dt}}{\mathrm{dx}} \\ & \Rightarrow\left(\mathrm{ex} \mathrm{e}^{\mathrm{e}-1}+\mathrm{e}^{\mathrm{x}}\right) \mathrm{dx}=\mathrm{dt} \\ & \therefore \int \frac{\left(\mathrm{x}^{\mathrm{e}-1}+\mathrm{e}^{\mathrm{x}-1}\right) \mathrm{dx}}{\mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}}=\frac{1}{\mathrm{e}} \int \frac{\mathrm{e} \cdot \mathrm{x}^{\mathrm{e}-1}+\mathrm{e}^{\mathrm{x}}}{\mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}} \cdot \mathrm{dx} \\ & =\frac{1}{\mathrm{e}} \int \frac{\mathrm{dt}}{\mathrm{t}}=\frac{1}{\mathrm{e}} \cdot \ln \mathrm{t}+\mathrm{c} \\ & =\frac{1}{\mathrm{e}} \cdot \ln \left(\mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}\right)+\mathrm{c}\end{aligned}
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