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Question

Mathematics Question on integral

If ∫dxx3(1+x6)23=f(x)(1+x6)13+C\int \frac{dx}{x^3 (1+x^6)^{\frac{2}{3}}} = f(x) (1+ x^{6})^{\frac{1}{3}} + C where C is a constant of integration, then f(x)f(x) is equal to :

A

−16x3- \frac{1}{6x^3}

B

−3x2- \frac{3}{x^2}

C

−12x2- \frac{1}{2x^2}

D

−12x3- \frac{1}{2x^3}

Answer

−12x3- \frac{1}{2x^3}

Explanation

Solution

∫dxx3(1+x6)23\int \frac{dx}{x^3(1+x^6)^{\frac{2}{3}}}

=∫dxx7(1+x6)23\int \frac{dx}{x^7(1+x^6)^{\frac{2}{3}}}
Let 1+1x6=t⇒−6x7 dx=dt1 + \frac{1}{x^6} = t \quad \Rightarrow \quad -6x^7 \, dx = dt
∴∴ I=−16∫dtt23I = -\frac{1}{6} \int \frac{dt}{t^{\frac{2}{3}}}
=$$-\frac{3}{6}t^{\frac{1}{3}} + C
=$$-\frac{1}{2}(1 + \frac{1}{x^6})^{\frac{1}{3}} + C
=$$-\frac{1}{2x^2}(1 + x^6)^{\frac{1}{3}} + C
∴∴ f(x)=−12x3f(x)=−\frac{1}{2x^3}