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

\frac{d}{d x} (\frac{4 x ^{2} - 16}{x - 2})

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

f (x) = \frac{1}{x}, (1, 1)

arccos (e^{9 x})

e^{- x + 1} \cdot x - 2 \cdot e^{- x + 1}

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