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)

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

\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{6}{x ^{3}}

\int \frac{3 x ^{2}}{4 5 - x ^{3}} d x

f (x) = 3 x^{2} - 3 x, at x = 3

x \to 0 lim (\frac{sinh ( x )}{x})

\frac{6 x ^{2} + 2 x + 4}{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