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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\int (- 2 x) d x

\frac{\partial}{\partial x} (x + y^{3})

\frac{\partial}{\partial x} (2 y x - 3 y + y^{2})

y e^{x y} d x + x e^{x y} d y = 0

\frac{d}{d x} (\frac{a}{b x ^{2}})

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