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

The integral of t/(sqrt(3t+4)) is 2/9 (1/3 (3t+4)^{3/2}-4sqrt(3t+4))+C

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integral of t/(sqrt(3t+4))

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

\frac{2}{9} (\frac{1}{3} (3 t + 4)^{\frac{3}{2}} - 4 3 t + 4) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 ( u ^{2} - 4 )}{9} d u

= \int \frac{2 ( u ^{2} - 4 )}{9} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int u^{2} - 4 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{2}{9} (\int u^{2} d u - \int 4 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 4 d u = 4 u

= \frac{2}{9} (\frac{u ^{3}}{3} - 4 u)

Substitute back

u = 3 t + 4

= \frac{2}{9} (\frac{( 3 t + 4 ) ^{3}}{3} - 4 3 t + 4)

Simplify

\frac{2}{9} (\frac{( 3 t + 4 ) ^{3}}{3} - 4 3 t + 4) : \frac{2}{9} (\frac{1}{3} (3 t + 4)^{\frac{3}{2}} - 4 3 t + 4)

= \frac{2}{9} (\frac{1}{3} (3 t + 4)^{\frac{3}{2}} - 4 3 t + 4)

Add a constant to the solution

= \frac{2}{9} (\frac{1}{3} (3 t + 4)^{\frac{3}{2}} - 4 3 t + 4) + C

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What is the integral of t/(sqrt(3t+4)) ?
The integral of t/(sqrt(3t+4)) is 2/9 (1/3 (3t+4)^{3/2}-4sqrt(3t+4))+C