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

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

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

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

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

Solution steps

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

Factor

\frac{x}{x ^{4} + 2 x ^{2} + 1}

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{2} (- \frac{1}{u})

Substitute back

u = x^{2} + 1

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

Simplify

\frac{1}{2} (- \frac{1}{x ^{2} + 1}) : - \frac{1}{2 ( x ^{2} + 1 )}

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (ln (x^{4} + y^{4}) - z)

\frac{\partial}{\partial x} (\frac{3}{x ^{3}})

\frac{d}{d x} (\frac{e ^{x}}{8 x ^{2} + 7})

f (x) = x (\frac{1}{x})

\int 15 d x

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