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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

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