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

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

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integral of (x^3)/((1+x^4)^2)

\int \frac{x ^{3}}{( 1 + x ^{4} ) ^{2}} d x

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

Solution steps

\int \frac{x ^{3}}{( 1 + x ^{4} ) ^{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 1 + x^{4}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} (y) = (y)^{2} \cdot e^{3 x}

\int \frac{( 1 )}{( cos ^{(4)} ( x ) )} d x

f (x) = 4 - 3 x

\frac{d}{d x} ((3 x + 8)^{2})

x \to 0 + lim (tan (5 x))

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