What is the integral from 0 to x of t/(t^2+1) ?

The integral from 0 to x of t/(t^2+1) is 1/2 ln|x^2+1|

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integral from 0 to x of t/(t^2+1)

\int_{0}^{x} \frac{t}{t ^{2} + 1} d t

\frac{1}{2} ln x^{2} + 1

Solution steps

\int_{0}^{x} \frac{t}{t ^{2} + 1} d t

Apply u-substitution

= \int_{1}^{x^{2} + 1} \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_{1}^{x^{2} + 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{x^{2} + 1}

Compute the boundaries:

ln x^{2} + 1

= \frac{1}{2} ln x^{2} + 1

What is the integral from 0 to x of t/(t^2+1) ?
The integral from 0 to x of t/(t^2+1) is 1/2 ln|x^2+1|