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

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

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

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

- ln 2 + 2 x^{2} + C

Solution steps

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

Take the constant out:

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

= - 4 \cdot \int \frac{x}{2 + 2 x ^{2}} d x

Apply u-substitution

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

= - 4 \cdot \int \frac{1}{4 u} d u

Take the constant out:

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

= - 4 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 4 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 + 2 x^{2}

= - 4 \cdot \frac{1}{4} ln 2 + 2 x^{2}

Simplify

- 4 \cdot \frac{1}{4} ln 2 + 2 x^{2} : - ln 2 + 2 x^{2}

= - ln 2 + 2 x^{2}

Add a constant to the solution

= - ln 2 + 2 x^{2} + C

\int \frac{csc ( y )}{1} d y

lo g_{10} (\frac{1}{x})

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

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

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