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

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

\int_{0}^{\frac{1}{8}} x arcsin (8 x) d x

\frac{d}{d x} ((3 x^{2} - 9)^{- 15})

y = (x^{2} - 2) (x^{3} - 3 x), (- 2, - 4)

\frac{d}{d x} (3 x^{2} - 4 x + 2)

y = x^{2} - 3 x + 3, (4, 7)

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