What is {dx}{x(xⁿ+1)} equal to?

Correct Answer is : 1nln⁡(xnxn+1)+c1/n≤ft({xⁿ}{xⁿ+1})+cn1ln(xn+1xn)+c

Correct Answer is : 1nln⁡(xnxn+1)+c1/n≤ft({xⁿ}{xⁿ+1})+cn1ln(xn+1xn)+c

Test Index

UPSC NDA I and II 2020 Math Solved Paper

Question : 73 of 120

Marks: +1, -0

What is

\int\frac{dx}{x(x^{n}+1)}

equal to?

Solution:

Concept:

Use substitution and partial fractions to simplify and integrate the given expression.

Explanation:

Multiply the numerator and denominator by

x^{n-1}

I = \int \frac{x^{n-1} \, dx}{x^n (x^n + 1)}

Substitute

t = x^n

. Then

dt = n x^{n-1} \, dx

, so

x^{n-1} dx = \frac{dt}{n}

Thus

I = \frac{1}{n} \int \frac{dt}{t(t+1)}

Use partial fractions:

\frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1}

Solve

1 = A(t+1) + Bt

. Put

t = 0 \Rightarrow A = 1

. Put

t = -1 \Rightarrow B = -1

Hence

\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}

Integrate:

I = \frac{1}{n} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt

= \frac{1}{n} \left( \ln|t| - \ln|t+1| \right) + c = \frac{1}{n} \ln \left| \frac{t}{t+1} \right| + c

Substitute back

t = x^n

I = \frac{1}{n} \ln \left( \frac{x^n}{x^n + 1} \right) + c

Answer:

\frac{1}{n}\ln\left(\frac{x^{n}}{x^{n}+1}\right)+c

(Option A).

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