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MHTCET 16 Apr 2026 Shift 2 Paper

Question : 128 of 150

Marks: +1, -0

\int \frac{x^{4}+1}{x^{6}+1} d x=

Solution:

Concept:

Factorize the denominator and split the numerator to simplify the integrand.

Explanation:

We have

x^6+1 = (x^2+1)(x^4 - x^2 + 1)

Write

x^4+1 = (x^4 - x^2 + 1) + x^2

Therefore,

\frac{x^4+1}{x^6+1} = \frac{1}{x^2+1} + \frac{x^2}{x^6+1}.

So,

\int \frac{x^4+1}{x^6+1}\,dx = \int \frac{dx}{1+x^2} + \int \frac{x^2}{1+x^6}\,dx.

Now

\int \frac{dx}{1+x^2} = \tan^{-1}x

For the second integral, let

x^3 = t

, so

3x^2\,dx = dt

Then,

\int \frac{x^2}{1+x^6}\,dx = \frac{1}{3}\int \frac{dt}{1+t^2} = \frac{1}{3}\tan^{-1}t.

Substitute

t = x^3

back,

\int \frac{x^2}{1+x^6}\,dx = \frac{1}{3}\tan^{-1}(x^3).

Thus the given integral equals

\tan^{-1}x + \frac{1}{3}\tan^{-1}(x^3) + c

Answer:

Option B:

\tan^{-1}x + \frac{1}{3}\tan^{-1}(x^3) + c

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