What is the integral of t/(3t^2+4) ?

The integral of t/(3t^2+4) is 1/6 ln|3t^2+4|+C

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integral of t/(3t^2+4)

\int \frac{t}{3 t ^{2} + 4} d t

\frac{1}{6} ln 3 t^{2} + 4 + C

Solution steps

\int \frac{t}{3 t ^{2} + 4} d t

Apply u-substitution

: \int \frac{1}{6 u} d u

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 3 t^{2} + 4

= \frac{1}{6} ln 3 t^{2} + 4

Add a constant to the solution

= \frac{1}{6} ln 3 t^{2} + 4 + C

x \to 0 lim (\frac{sin ( x )}{sin ( 2 x )})

\int \frac{( x ^{2} + 1 )}{x ^{3} + 1} d x

n = 1 \sum \infty \frac{1}{( 4 n ^{2} - 1 )}

\int_{0}^{\frac{π}{2}} cos (t) d t

y^{^{'''}} + 3 y^{^{''}} - 4 y^{^{'}} - 12 = 0

What is the integral of t/(3t^2+4) ?
The integral of t/(3t^2+4) is 1/6 ln|3t^2+4|+C