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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int 2 (u^{2} - 3) d u

= \int 2 (u^{2} - 3) d u

Take the constant out:

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

= 2 \cdot \int u^{2} - 3 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

= 2 (\int u^{2} d u - \int 3 d u)

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

\int 3 d u = 3 u

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

Substitute back

u = t + 3

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

Simplify

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

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

Add a constant to the solution

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

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