What is the integral of t/(sqrt(1+3t)) ?

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

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integral of t/(sqrt(1+3t))

\int \frac{t}{1 + 3 t} d t

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

Solution steps

\int \frac{t}{1 + 3 t} d t

Apply u-substitution

: \int \frac{2 ( u ^{2} - 1 )}{9} d u

= \int \frac{2 ( u ^{2} - 1 )}{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} - 1 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 1 d u)

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

\int 1 d u = u

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

Substitute back

u = 1 + 3 t

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

Simplify

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

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

Add a constant to the solution

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

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