What is the integral of t/(2t^2+3) ?

The integral of t/(2t^2+3) is 1/4 ln|2t^2+3|+C

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integral of t/(2t^2+3)

\int \frac{t}{2 t ^{2} + 3} d t

\frac{1}{4} ln 2 t^{2} + 3 + C

Solution steps

\int \frac{t}{2 t ^{2} + 3} d t

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 t^{2} + 3

= \frac{1}{4} ln 2 t^{2} + 3

Add a constant to the solution

= \frac{1}{4} ln 2 t^{2} + 3 + C

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What is the integral of t/(2t^2+3) ?
The integral of t/(2t^2+3) is 1/4 ln|2t^2+3|+C