What is the integral of (x^7)/((1+x^4)^2) ?

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

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integral of (x^7)/((1+x^4)^2)

\int \frac{x ^{7}}{( 1 + x ^{4} ) ^{2}} d x

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

Solution steps

\int \frac{x ^{7}}{( 1 + x ^{4} ) ^{2}} d x

Apply u-substitution

: \int \frac{u - 1}{4 u ^{2}} d u

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

Expand

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

Substitute back

u = 1 + x^{4}

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

Simplify

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

Add a constant to the solution

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

(\frac{( e ^{6} )}{( e ^{4} )})^{- 1}

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 4})

x \frac{d v}{d x} = \frac{1 - 4 v ^{2}}{3 v}

(x^{2} + y^{2} - 7) d x = (y + x y) d y, y (0) = 1

\frac{d}{d x} (lo g_{3} (\frac{3 x ^{2}}{x}))

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