What is the integral of (1000x)/(x^2+e) ?

The integral of (1000x)/(x^2+e) is 500ln|x^2+e|+C

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integral of (1000x)/(x^2+e)

\int \frac{1000 x}{x ^{2} + e} d x

500 ln x^{2} + e + C

Solution steps

\int \frac{1000 x}{x ^{2} + e} d x

Take the constant out:

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

= 1000 \cdot \int \frac{x}{x ^{2} + e} d x

Apply u-substitution

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

= 1000 \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

= 1000 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 1000 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + e

= 1000 \cdot \frac{1}{2} ln x^{2} + e

Simplify

1000 \cdot \frac{1}{2} ln x^{2} + e : 500 ln x^{2} + e

= 500 ln x^{2} + e

Add a constant to the solution

= 500 ln x^{2} + e + C

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What is the integral of (1000x)/(x^2+e) ?
The integral of (1000x)/(x^2+e) is 500ln|x^2+e|+C