What is the integral of (x^3)/((a+bx^2)^3) ?

The integral of (x^3)/((a+bx^2)^3) is (-2bx^2-a)/(4b^2(bx^2+a)^2)+C

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integral of (x^3)/((a+bx^2)^3)

\int \frac{x ^{3}}{( a + b x ^{2} ) ^{3}} d x

\frac{- 2 b x ^{2} - a}{4 b ^{2} ( b x ^{2} + a ) ^{2}} + C

Solution steps

\int \frac{x ^{3}}{( a + b x ^{2} ) ^{3}} d x

Apply u-substitution

: \int \frac{u - a}{2 b ^{2} u ^{3}} d u

= \int \frac{u - a}{2 b ^{2} u ^{3}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2 b ^{2}} \cdot \int \frac{u - a}{u ^{3}} d u

Expand

\frac{u - a}{u ^{3}} : \frac{1}{u ^{2}} - \frac{a}{u ^{3}}

= \frac{1}{2 b ^{2}} \cdot \int \frac{1}{u ^{2}} - \frac{a}{u ^{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2 b ^{2}} (\int \frac{1}{u ^{2}} d u - \int \frac{a}{u ^{3}} d u)

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{a}{u ^{3}} d u = - \frac{a}{2 u ^{2}}

= \frac{1}{2 b ^{2}} (- \frac{1}{u} - (- \frac{a}{2 u ^{2}}))

Substitute back

u = a + b x^{2}

= \frac{1}{2 b ^{2}} (- \frac{1}{a + b x ^{2}} - (- \frac{a}{2 ( a + b x ^{2} ) ^{2}}))

Simplify

\frac{1}{2 b ^{2}} (- \frac{1}{a + b x ^{2}} - (- \frac{a}{2 ( a + b x ^{2} ) ^{2}})) : \frac{- 2 b x ^{2} - a}{4 b ^{2} ( b x ^{2} + a ) ^{2}}

= \frac{- 2 b x ^{2} - a}{4 b ^{2} ( b x ^{2} + a ) ^{2}}

Add a constant to the solution

= \frac{- 2 b x ^{2} - a}{4 b ^{2} ( b x ^{2} + a ) ^{2}} + C

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What is the integral of (x^3)/((a+bx^2)^3) ?
The integral of (x^3)/((a+bx^2)^3) is (-2bx^2-a)/(4b^2(bx^2+a)^2)+C