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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{8} ln ∣ u ∣

Substitute back

u = 4 x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int - e^{y} y^{2} d y

x + 2

x - 2

\int - ln (x^{2} - x + 1) d x

\int \frac{1}{( x ^{2} + 2 x + 2 ) ^{\frac{3}{2}}} d x

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