What is the integral of (x^5)/((3-x^2)^4) ?

The integral of (x^5)/((3-x^2)^4) is 1/2 ((x^4)/(3(-x^2+3)^3)-(2x^2-3)/(3(-x^2+3)^2))+C

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integral of (x^5)/((3-x^2)^4)

\int \frac{x ^{5}}{( 3 - x ^{2} ) ^{4}} d x

\frac{1}{2} (\frac{x ^{4}}{3 ( - x ^{2} + 3 ) ^{3}} - \frac{2 x ^{2} - 3}{3 ( - x ^{2} + 3 ) ^{2}}) + C

Solution steps

\int \frac{x ^{5}}{( 3 - x ^{2} ) ^{4}} d x

Apply u-substitution

: \int \frac{u ^{2}}{2 ( - u + 3 ) ^{4}} d u

= \int \frac{u ^{2}}{2 ( - u + 3 ) ^{4}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u ^{2}}{( - u + 3 ) ^{4}} d u

Apply Integration By Parts

: \frac{u ^{2}}{3 ( - u + 3 ) ^{3}} - \int \frac{2 u}{3 ( - u + 3 ) ^{3}} d u

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

\int \frac{2 u}{3 ( - u + 3 ) ^{3}} d u = \frac{2 u - 3}{3 ( - u + 3 ) ^{2}}

= \frac{1}{2} (\frac{u ^{2}}{3 ( - u + 3 ) ^{3}} - \frac{2 u - 3}{3 ( - u + 3 ) ^{2}})

Substitute back

u = x^{2}

= \frac{1}{2} (\frac{( x ^{2} ) ^{2}}{3 ( - x ^{2} + 3 ) ^{3}} - \frac{2 x ^{2} - 3}{3 ( - x ^{2} + 3 ) ^{2}})

(x^{2})^{2} = x^{4}

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

Add a constant to the solution

= \frac{1}{2} (\frac{x ^{4}}{3 ( - x ^{2} + 3 ) ^{3}} - \frac{2 x ^{2} - 3}{3 ( - x ^{2} + 3 ) ^{2}}) + C

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What is the integral of (x^5)/((3-x^2)^4) ?
The integral of (x^5)/((3-x^2)^4) is 1/2 ((x^4)/(3(-x^2+3)^3)-(2x^2-3)/(3(-x^2+3)^2))+C