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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2020 · Mathematics · Indefinite Integration

What is ∫ (d x)/(x(x^n+1)) equal to?

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

Options

  1. A1nln⁡(xnxn+1)+c\frac{1}{n} \ln \left(\frac{x^n}{x^n+1}\right)+c
  2. Bln⁡(xn+1xn)+c\ln \left(\frac{x^n+1}{x^n}\right)+c
  3. Cln⁡(xnxn+1)+c\ln \left(\frac{x^n}{x^n+1}\right)+c
  4. D1nln⁡(xn+1xn)+c\frac{1}{n} \ln \left(\frac{x^n+1}{x^n}\right)+c

Correct answer

A. 1nln⁡(xnxn+1)+c\frac{1}{n} \ln \left(\frac{x^n}{x^n+1}\right)+c

Step-by-step solution

∫dxx(xn+1)=∫xn−1xn(xn+1)⋅dx\int \frac{d x}{x\left(x^n+1\right)}=\int \frac{x^{n-1}}{x^n\left(x^n+1\right)} \cdot d x Let z=xn⇒dz=nxn−1dx∴∫dxx(xn+1)=∫xn−1xn(xn+1)⋅dx=∫1z(z+1)⋅dznz=x^n \Rightarrow d z=n x^{n-1} d x\therefore \int \frac{d x}{x\left(x^n+1\right)}=\int \frac{x^{n-1}}{x^n\left(x^n+1\right)} \cdot d x=\int \frac{1}{z(z+1)} \cdot \frac{d z}{n}=1n[∫1z⋅dz−∫dzz+1]=1n[ln⁡z−ln⁡(z+1)]+C where C= constant \begin{aligned} & =\frac{1}{n}\left[\int \frac{1}{z} \cdot d z-\int \frac{d z}{z+1}\right] \\ & =\frac{1}{n}[\ln z-\ln (z+1)]+C \quad \text{ where } C=\text{ constant }\end{aligned}=1n⋅ln⁡=\frac{1}{n} \cdot \ln\left(\frac{z}{z+1}\right)+C=\frac{1}{n} \ln \left(\frac{x^n}{x^n+1}\right)+C
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