What is the integral of 3/2 sqrt(t+1) ?

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

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integral of 3/2 sqrt(t+1)

\int \frac{3}{2} t + 1 d t

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

Solution steps

\int \frac{3}{2} t + 1 d t

Take the constant out:

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

= \frac{3}{2} \cdot \int t + 1 d t

Apply u-substitution

: \int u d u

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{3}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = t + 1

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

Simplify

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

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

Add a constant to the solution

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

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