What is the integral of (3x)/(x^2+1) ?

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

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integral of (3x)/(x^2+1)

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

\frac{3}{2} ln x^{2} + 1 + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 3 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 1

= 3 \cdot \frac{1}{2} ln x^{2} + 1

Simplify

3 \cdot \frac{1}{2} ln x^{2} + 1 : \frac{3}{2} ln x^{2} + 1

= \frac{3}{2} ln x^{2} + 1

Add a constant to the solution

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

\int \frac{2 x - 3}{x ^{2} - 3 x + 2} d x

\frac{d}{d x} (e^{3 x - 2})

\int cot^{3} (5 x) d x

f (x) = 8 x sin (π x)

4 t^{- \frac{1}{8}} + 4 ln (t) + ln (5)

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