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

The integral of x/(1+4x^2) is 1/8 ln|1+4x^2|+C

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

u = 1 + 4 x^{2}

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

Add a constant to the solution

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

f (x) = \frac{2 x}{3}

y^{^{'}} = x (8 - y)

\int cos (2 - 3 x) d x

\frac{d}{d x} (x + \frac{c}{x})

\frac{e ^{t^{2}} - 1}{t}

What is the integral of x/(1+4x^2) ?
The integral of x/(1+4x^2) is 1/8 ln|1+4x^2|+C