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

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

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

integral

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{8 u} d u

= 2 \cdot \int \frac{1}{8 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{8} ln ∣ u ∣

Substitute back

u = 1 + 4 x^{2}

= 2 \cdot \frac{1}{8} ln 1 + 4 x^{2}

Simplify

2 \cdot \frac{1}{8} ln 1 + 4 x^{2} : \frac{1}{4} ln 1 + 4 x^{2}

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

Add a constant to the solution

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

\frac{1}{( x ^{3} + x )}

(e^{- t})^{2}

2 t + 5

5 x^{2}

sec^{2} (x) tan^{2} (x)

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