What is the integral from 0 to x of t/(9+t^2) ?

The integral from 0 to x of t/(9+t^2) is 1/2 ln|9+x^2|-ln(3)

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integral from 0 to x of t/(9+t^2)

\int_{0}^{x} \frac{t}{9 + t ^{2}} d t

\frac{1}{2} ln 9 + x^{2} - ln (3)

Solution steps

\int_{0}^{x} \frac{t}{9 + t ^{2}} d t

Apply u-substitution

: \int_{9}^{9 + x^{2}} \frac{1}{2 u} d u

= \int_{9}^{9 + x^{2}} \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_{9}^{9 + x^{2}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{9}^{9 + x^{2}}

Compute the boundaries:

ln 9 + x^{2} - 2 ln (3)

= \frac{1}{2} (ln 9 + x^{2} - 2 ln (3))

Simplify

= \frac{1}{2} ln 9 + x^{2} - ln (3)

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What is the integral from 0 to x of t/(9+t^2) ?
The integral from 0 to x of t/(9+t^2) is 1/2 ln|9+x^2|-ln(3)