What is the integral of (x^7)/((4-x^4)^2) ?

The integral of (x^7)/((4-x^4)^2) is 1/4 ln|4-x^4|+1/(4-x^4)+C

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integral of (x^7)/((4-x^4)^2)

\int \frac{x ^{7}}{( 4 - x ^{4} ) ^{2}} d x

\frac{1}{4} ln 4 - x^{4} + \frac{1}{4 - x ^{4}} + C

Solution steps

\int \frac{x ^{7}}{( 4 - x ^{4} ) ^{2}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{- u + 4}{u ^{2}} : - \frac{1}{u} + \frac{4}{u ^{2}}

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{4}{u ^{2}} d u = - \frac{4}{u}

= - \frac{1}{4} (- ln ∣ u ∣ - \frac{4}{u})

Substitute back

u = 4 - x^{4}

= - \frac{1}{4} (- ln 4 - x^{4} - \frac{4}{4 - x ^{4}})

Simplify

- \frac{1}{4} (- ln 4 - x^{4} - \frac{4}{4 - x ^{4}}) : \frac{1}{4} ln 4 - x^{4} + \frac{1}{4 - x ^{4}}

= \frac{1}{4} ln 4 - x^{4} + \frac{1}{4 - x ^{4}}

Add a constant to the solution

= \frac{1}{4} ln 4 - x^{4} + \frac{1}{4 - x ^{4}} + C

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What is the integral of (x^7)/((4-x^4)^2) ?
The integral of (x^7)/((4-x^4)^2) is 1/4 ln|4-x^4|+1/(4-x^4)+C