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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= - 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 + t^{2}

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

Simplify

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

= - 1000 ln 1 + t^{2}

Add a constant to the solution

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

\frac{d}{d x} (4 x^{2} + x)

x^{2} - 6 x

x \to - 2 - lim (- 1)

3 (x^{2} + 3 x + 1)^{2} (2 x + 3)

2 x^{7}

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