What is the integral of t/(t^2+9) ?

The integral of t/(t^2+9) is 1/2 ln|t^2+9|+C

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integral of t/(t^2+9)

\int \frac{t}{t ^{2} + 9} d t

\frac{1}{2} ln t^{2} + 9 + C

Solution steps

\int \frac{t}{t ^{2} + 9} 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 = t^{2} + 9

= \frac{1}{2} ln t^{2} + 9

Add a constant to the solution

= \frac{1}{2} ln t^{2} + 9 + C

\int \frac{y ^{3} + 1}{y} d y

\int (x + y)^{- 2} d x

\int \frac{x ^{2} + 1}{x + 1} d x

\int e^{5 x} cos (\frac{x}{3}) d x

e^{(- x^{2})}

What is the integral of t/(t^2+9) ?
The integral of t/(t^2+9) is 1/2 ln|t^2+9|+C