What is the integral of-(2000x)/((1+x^2)) ?

The integral of-(2000x)/((1+x^2)) is -1000ln|1+x^2|+C

integral of-(2000x)/((1+x^2))

\int - \frac{2000 x}{( 1 + x ^{2} )} d x

- 1000 ln 1 + x^{2} + C

Solution steps

\int - \frac{2000 x}{1 + x ^{2}} d x

Take the constant out:

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

= - 2000 \cdot \int \frac{x}{1 + x ^{2}} d x

Apply u-substitution

= - 2000 \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

= - 2000 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2000 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + x^{2}

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

Simplify

- 2000 \cdot \frac{1}{2} ln 1 + x^{2} : - 1000 ln 1 + x^{2}

= - 1000 ln 1 + x^{2}

Add a constant to the solution

= - 1000 ln 1 + x^{2} + C

t an g e n t x^{2}, at x = 2

\frac{\partial}{\partial x} (3 x y^{2} + 20 x^{2} y^{3})

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (\frac{π}{4}) = 2, y^{^{'}} (\frac{π}{4}) = - 2

\int \frac{3 x ^{2} + 17 x + 32}{x ^{3} + 8 x ^{2} + 16 x} d x

\int (x + 1) 3 x + 2 d x

What is the integral of-(2000x)/((1+x^2)) ?
The integral of-(2000x)/((1+x^2)) is -1000ln|1+x^2|+C