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

The integral of x/((1+x^2)sqrt(1-x^4)) is -1/(tan(1/2 arcsin(x^2))+1)+C

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integral of x/((1+x^2)sqrt(1-x^4))

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

- \frac{1}{tan ( \frac{1}{2} arcsin ( x ^{2} ) ) + 1} + C

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{1 + sin ( v )} d v

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

Apply u-substitution

: \int \frac{2}{1 + w ^{2} + 2 w} d w

= \frac{1}{2} \cdot \int \frac{2}{1 + w ^{2} + 2 w} d w

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{1 + w ^{2} + 2 w} d w

Complete the square

1 + w^{2} + 2 w : (w + 1)^{2}

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

Apply u-substitution

: \int \frac{1}{t ^{2}} d t

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{t ^{2}} d t

Apply the Power Rule

: - \frac{1}{t}

= \frac{1}{2} \cdot 2 (- \frac{1}{t})

Substitute back

= \frac{1}{2} \cdot 2 - \frac{1}{tan ( \frac{a r c s i n ( x ^{2} )}{2} ) + 1}

Simplify

\frac{1}{2} \cdot 2 - \frac{1}{tan ( \frac{a r c s i n ( x ^{2} )}{2} ) + 1} : - \frac{1}{tan ( \frac{1}{2} arcsin ( x ^{2} ) ) + 1}

= - \frac{1}{tan ( \frac{1}{2} arcsin ( x ^{2} ) ) + 1}

Add a constant to the solution

= - \frac{1}{tan ( \frac{1}{2} arcsin ( x ^{2} ) ) + 1} + C

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