What is the integral of (9x)/(x^2+1) ?

The integral of (9x)/(x^2+1) is 9/2 ln|x^2+1|+C

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integral of (9x)/(x^2+1)

\int \frac{9 x}{x ^{2} + 1} d x

\frac{9}{2} ln x^{2} + 1 + C

Solution steps

\int \frac{9 x}{x ^{2} + 1} d x

Take the constant out:

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

= 9 \cdot \int \frac{x}{x ^{2} + 1} d x

Apply u-substitution

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

= 9 \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

= 9 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 1

= 9 \cdot \frac{1}{2} ln x^{2} + 1

Simplify

9 \cdot \frac{1}{2} ln x^{2} + 1 : \frac{9}{2} ln x^{2} + 1

= \frac{9}{2} ln x^{2} + 1

Add a constant to the solution

= \frac{9}{2} ln x^{2} + 1 + C

f (x) = 6 x, at x = 1

(16 y^{2} - x y) d x + x^{2} d y = 0

\frac{d}{d x} (2^{x} tan (3 x))

\frac{\partial}{\partial z} (x y + 2 z)

What is the integral of (9x)/(x^2+1) ?
The integral of (9x)/(x^2+1) is 9/2 ln|x^2+1|+C