CUET Mathematics 2024 - x^n+1-x d x= | PYQs + Solutions

CUET Mathematics 2024

Calculus

Integrals

Medium

$\int \dfrac{\pi}{x^{n+1}-x} d x=$

\frac{\pi}{n} \log _{e}\left|\frac{x^{n}-1}{x^{n}}\right|+C

\quad \log _{e}\left|\frac{x^{n}+1}{x^{n}-1}\right|+C

\frac{\pi}{n} \log _{e}\left|\frac{x^{n}+1}{x^{n}}\right|+C

\pi \log _{e}\left|\frac{x^{n}}{x^{n}-1}\right|+C

✅ Correct Option: 1

We need to evaluate

\int \dfrac{\pi}{x^{n+1}-x} dx

First, let's rewrite the denominator:

x^{n+1} - x = x(x^n - 1)

So our integral becomes:

\int \dfrac{\pi}{x(x^n-1)} dx

Using partial fractions:

\dfrac{\pi}{x(x^n-1)} = \dfrac{A}{x} + \dfrac{B}{x^n-1}

Multiplying all terms by

x(x^n-1)

\pi = A(x^n-1) + Bx

When

x = 0

\pi = A(-1)

, so

A = -\pi

When

x = 1

\pi = B

, so

B = \pi

Therefore:

\dfrac{\pi}{x^{n+1}-x} = \dfrac{-\pi}{x} + \dfrac{\pi}{x^n-1}

Integrating each term:

\int \dfrac{\pi}{x^{n+1}-x}dx = \int \dfrac{-\pi}{x}dx + \int \dfrac{\pi}{x^n-1}dx

= -\pi\ln|x| + \pi\int \dfrac{1}{x^n-1}dx

For the second integral, use substitution

t = x^n

dt = nx^{n-1}dx

→

dx = \dfrac{dt}{nx^{n-1}}

\int \dfrac{1}{x^n-1}dx = \int \dfrac{1}{t-1} \cdot \dfrac{dt}{nx^{n-1}}

Since

x^{n-1} = \dfrac{t}{x}

, we have:

\int \dfrac{1}{x^n-1}dx = \dfrac{1}{n}\int \dfrac{x}{t(t-1)}dt

Let's try direct substitution with

t = x^n

\int \dfrac{\pi}{x(x^n-1)}dx = \dfrac{\pi}{n} \int \dfrac{dt}{t(t-1)}

= \dfrac{\pi}{n} \int \left(\dfrac{1}{t-1} - \dfrac{1}{t}\right)dt

= \dfrac{\pi}{n}[\ln|t-1| - \ln|t|] + C

= \dfrac{\pi}{n}\ln\left|\dfrac{t-1}{t}\right| + C

Substituting back

t = x^n

\int \dfrac{\pi}{x^{n+1}-x}dx=\boxed{\dfrac{\pi}{n}\ln\left|\dfrac{x^n-1}{x^n}\right| + C}