If ^4+1/x^6+1\,dx = A^{-1} x + B^{-1} x³ + C, then (A, B) =

Correct Answer is : (1,13)≤ft(1, 1/3 )(1,31)

Correct Answer is : (1,13)≤ft(1, 1/3 )(1,31)

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TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 72 of 160

Marks: +1, -0

\int\frac{x^4+1}{x^6+1}\,dx = A\tan^{-1} x

+ B\tan^{-1} x^3 + C,

(A, B) =

Solution:

Consider the integral.

\int\frac{x^4+1}{x^6+1}\,dx = A\tan^{-1} x

+ B\tan^{-1} x^3 + C

Let

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

The above integral is solved as,

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

=\int\frac{x^4+1+x^2-x^2}{(x^2)^3+1^3}\,dx

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

+\int\frac{x^2}{((x^2)^3+1^3)(x^4+1-x^2)}\,dx

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

Let

x^3 = t

3 x^2 dx = dt

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

=\tan^{-1} x+ \frac{1}{3} \tan^{-1} t+c

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

Compare with the original equation.

A=1

B= \frac{1}{3}

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