What is the integral of-(4000x)/((1+x^2)) ?

The integral of-(4000x)/((1+x^2)) is -2000ln|1+x^2|+C

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integral of-(4000x)/((1+x^2))

\int - \frac{4000 x}{( 1 + x ^{2} )} d x

- 2000 ln 1 + x^{2} + C

Solution steps

\int - \frac{4000 x}{1 + x ^{2}} d x

Take the constant out:

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

= - 4000 \cdot \int \frac{x}{1 + x ^{2}} d x

Apply u-substitution

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

= - 4000 \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

= - 4000 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 4000 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + x^{2}

= - 4000 \cdot \frac{1}{2} ln 1 + x^{2}

Simplify

- 4000 \cdot \frac{1}{2} ln 1 + x^{2} : - 2000 ln 1 + x^{2}

= - 2000 ln 1 + x^{2}

Add a constant to the solution

= - 2000 ln 1 + x^{2} + C

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

\frac{d y}{d x} + 10 tan (10 x) y = 5 cos (10 x)

y^{^{'}} = y^{2} sin (t)

x \to 0 lim (x csc (4 x))

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What is the integral of-(4000x)/((1+x^2)) ?
The integral of-(4000x)/((1+x^2)) is -2000ln|1+x^2|+C