What is the integral of (2u)/(4u^2+3) ?

The integral of (2u)/(4u^2+3) is 1/4 ln|4u^2+3|+C

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integral of (2u)/(4u^2+3)

\int \frac{2 u}{4 u ^{2} + 3} d u

\frac{1}{4} ln 4 u^{2} + 3 + C

Solution steps

\int \frac{2 u}{4 u ^{2} + 3} d u

Take the constant out:

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

= 2 \cdot \int \frac{u}{4 u ^{2} + 3} d u

Apply u-substitution

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

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}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{8} ln ∣ v ∣

Substitute back

v = 4 u^{2} + 3

= 2 \cdot \frac{1}{8} ln 4 u^{2} + 3

Simplify

2 \cdot \frac{1}{8} ln 4 u^{2} + 3 : \frac{1}{4} ln 4 u^{2} + 3

= \frac{1}{4} ln 4 u^{2} + 3

Add a constant to the solution

= \frac{1}{4} ln 4 u^{2} + 3 + C

What is the integral of (2u)/(4u^2+3) ?
The integral of (2u)/(4u^2+3) is 1/4 ln|4u^2+3|+C