What is the integral of (x^3)/((1+x^8)^2) ?

The integral of (x^3)/((1+x^8)^2) is 1/8 (arctan(x^4)+1/2 sin(2arctan(x^4)))+C

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integral of (x^3)/((1+x^8)^2)

\int \frac{x ^{3}}{( 1 + x ^{8} ) ^{2}} d x

\frac{1}{8} (arctan (x^{4}) + \frac{1}{2} sin (2 arctan (x^{4}))) + C

Solution steps

\int \frac{x ^{3}}{( 1 + x ^{8} ) ^{2}} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= \frac{1}{4} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= \frac{1}{4} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

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} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= \frac{1}{4} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= \frac{1}{4} \cdot \frac{1}{2} (arctan (x^{4}) + \frac{1}{2} sin (2 arctan (x^{4})))

Simplify

\frac{1}{4} \cdot \frac{1}{2} (arctan (x^{4}) + \frac{1}{2} sin (2 arctan (x^{4}))) : \frac{1}{8} (arctan (x^{4}) + \frac{1}{2} sin (2 arctan (x^{4})))

= \frac{1}{8} (arctan (x^{4}) + \frac{1}{2} sin (2 arctan (x^{4})))

Add a constant to the solution

= \frac{1}{8} (arctan (x^{4}) + \frac{1}{2} sin (2 arctan (x^{4}))) + C

\int \frac{3}{x ( ln ( x ) ) ^{4}} d x

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

\int 2 (x^{2} + x + 2) e^{- 2 x} d x

cos^{3} (x)

What is the integral of (x^3)/((1+x^8)^2) ?
The integral of (x^3)/((1+x^8)^2) is 1/8 (arctan(x^4)+1/2 sin(2arctan(x^4)))+C