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

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

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

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

- \frac{1}{2} ln 1 + t^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= - \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + t^{2}

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

Add a constant to the solution

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

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