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

The integral of x/((4+x^4)^2) is 1/32 (arctan((x^2)/2)+1/2 sin(2arctan((x^2)/2)))+C

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

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

\frac{1}{32} (arctan (\frac{x ^{2}}{2}) + \frac{1}{2} sin (2 arctan (\frac{x ^{2}}{2}))) + C

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{8 ( v ^{2} + 1 ) ^{2}} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{8} \cdot \int \frac{1}{( v ^{2} + 1 ) ^{2}} d v

Apply Trigonometric Substitution

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

= \frac{1}{2} \cdot \frac{1}{8} \cdot \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{1}{8} \cdot \int \frac{1 + cos ( 2 w )}{2} d w

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{8} \cdot \frac{1}{2} \cdot \int 1 + cos (2 w) d w

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}{2} \cdot \frac{1}{8} \cdot \frac{1}{2} (\int 1 d w + \int cos (2 w) d w)

\int 1 d w = w

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

= \frac{1}{2} \cdot \frac{1}{8} \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w))

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{8} \cdot \frac{1}{2} (arctan (\frac{x ^{2}}{2}) + \frac{1}{2} sin (2 arctan (\frac{x ^{2}}{2}))) : \frac{1}{32} (arctan (\frac{x ^{2}}{2}) + \frac{1}{2} sin (2 arctan (\frac{x ^{2}}{2})))

= \frac{1}{32} (arctan (\frac{x ^{2}}{2}) + \frac{1}{2} sin (2 arctan (\frac{x ^{2}}{2})))

Add a constant to the solution

= \frac{1}{32} (arctan (\frac{x ^{2}}{2}) + \frac{1}{2} sin (2 arctan (\frac{x ^{2}}{2}))) + C

\frac{d}{d x} (- arctan (\frac{x}{y}))

\frac{d}{d x} (sin (x^{2} - x))

\int tan (x) sec^{6} (x) d x

y = (1 + 3 x)^{8}, (0, 1)

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

What is the integral of x/((4+x^4)^2) ?
The integral of x/((4+x^4)^2) is 1/32 (arctan((x^2)/2)+1/2 sin(2arctan((x^2)/2)))+C