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$\int \frac{dx}{x^{2} {(x^{4} + 1)}^{3 / 4}} = A {(\frac{x^{4} + 1}{x^{4}})}^{B} + C$

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A=−1,B=14

A=1,B=−14

A=12,B=1

A=−12,B=−1

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detailed solution

Correct option is A

I=∫dxx2x4+13/4=∫dxx2⋅x31+1x43/4Put⁡1+x−4=t⇒−4x5dx=dt⇒−14∫dtt3/4=14∫t−3/4dt=−14⋅t1/41/4+C=−1+1x41/4+C=−1+x4x41/4+C Hence, A=−1,B=14

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∫dxx2x4+13/4=Ax4+1x4B+C

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$\int \frac{dx}{x^{2} {(x^{4} + 1)}^{3 / 4}} = A {(\frac{x^{4} + 1}{x^{4}})}^{B} + C$