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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2021 · Mathematics · Indefinite Integration

What is ∫ (d x)/(x(x^2+1)) equal to?

What is ∫dxx(x2+1)\int \frac{d x}{x\left(x^2+1\right)} equal to?

Options

  1. A12ln⁡(x2x2+1)+C\frac{1}{2} \ln \left(\frac{x^2}{x^2+1}\right)+C
  2. Bln⁡(x2x2+1)+C\ln \left(\frac{x^2}{x^2+1}\right)+C
  3. C32ln⁡(x2x2+1)+C\frac{3}{2} \ln \left(\frac{x^2}{x^2+1}\right)+C
  4. D12ln⁡(x2+1x2)+C\frac{1}{2} \ln \left(\frac{x^2+1}{x^2}\right)+C

Correct answer

A. 12ln⁡(x2x2+1)+C\frac{1}{2} \ln \left(\frac{x^2}{x^2+1}\right)+C

Step-by-step solution

I=∫dxx(x2+1)=∫dxx(x2+1)Let x2+1=t⇒x2=t−12xds=dt=12∫dtt(t−1)=−12∫[1t−1(t−1)]dt=−12[∫1tdt−∫1(t−1)dt]=−12[log⁡(t)−log⁡(t−1)]+C=12[log⁡(t−1)−log⁡t]+C\begin{aligned} & \mathrm{I}=\int \frac{d x}{x\left(x^2+1\right)} \\ & =\int \frac{d x}{x\left(x^2+1\right)} \\ & \text{Let } x^2+1=t \Rightarrow x^2=t-1 \\ & 2 x d s=d t \\ & =\frac{1}{2} \int \frac{d t}{t(t-1)} \\ & =\frac{-1}{2} \int\left[\frac{1}{t}-\frac{1}{(t-1)}\right] d t \\ & =\frac{-1}{2}\left[\int \frac{1}{t} d t-\int \frac{1}{(t-1)} d t\right] \\ & =\frac{-1}{2}[\log (t)-\log (t-1)]+C \\ & =\frac{1}{2}[\log (t-1)-\log t]+C\end{aligned}=12log⁡(t−1t)+C=12log⁡(x2x2+1)+C\begin{aligned} & =\frac{1}{2} \log \left(\frac{t-1}{t}\right)+C \\ & =\frac{1}{2} \log \left(\frac{x^2}{x^2+1}\right)+C\end{aligned}$
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