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Worked solution

Use the derivative to find the integral

Find ddx(x31+x4)\dfrac{d}{dx}\left(\dfrac{x^3}{1+x^4}\right) and hence ∫4x2(3−x4)(1+x4)2 dx\displaystyle\int\dfrac{4x^2(3-x^4)}{(1+x^4)^2}\,dx.

CJC · Discussion · Question 1

Answer

x2(3−x4)(1+x4)2\dfrac{x^2(3-x^4)}{(1+x^4)^2}; hence 4x31+x4+C\dfrac{4x^3}{1+x^4}+C.

Differentiate

Apply the quotient rule:

ddx(x31+x4)=(1+x4)(3x2)−x3(4x3)(1+x4)2\dfrac{d}{dx}\left(\dfrac{x^3}{1+x^4}\right)=\dfrac{(1+x^4)(3x^2)-x^3(4x^3)}{(1+x^4)^2}

=3x2+3x6−4x6(1+x4)2=x2(3−x4)(1+x4)2=\dfrac{3x^2+3x^6-4x^6}{(1+x^4)^2}=\dfrac{x^2(3-x^4)}{(1+x^4)^2}.

How did this part go?

Hence integrate

The integrand is four times the derivative just obtained:

∫4x2(3−x4)(1+x4)2 dx=4∫x2(3−x4)(1+x4)2 dx=4x31+x4+C\displaystyle\int\dfrac{4x^2(3-x^4)}{(1+x^4)^2}\,dx=4\int\dfrac{x^2(3-x^4)}{(1+x^4)^2}\,dx=\dfrac{4x^3}{1+x^4}+C.

How did this part go?

Common mistakes

“Hence” asks you to use the earlier result. Keep the factor 4 and the constant of integration.

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