What is the integral of (1100x)/(x^2+80) ?

The integral of (1100x)/(x^2+80) is 550ln|x^2+80|+C

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integral of (1100x)/(x^2+80)

\int \frac{1100 x}{x ^{2} + 80} d x

550 ln x^{2} + 80 + C

Solution steps

\int \frac{1100 x}{x ^{2} + 80} d x

Take the constant out:

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

= 1100 \cdot \int \frac{x}{x ^{2} + 80} d x

Apply u-substitution

= 1100 \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

= 1100 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 1100 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 80

= 1100 \cdot \frac{1}{2} ln x^{2} + 80

Simplify

1100 \cdot \frac{1}{2} ln x^{2} + 80 : 550 ln x^{2} + 80

= 550 ln x^{2} + 80

Add a constant to the solution

= 550 ln x^{2} + 80 + C

What is the integral of (1100x)/(x^2+80) ?
The integral of (1100x)/(x^2+80) is 550ln|x^2+80|+C