What is the integral of 9/2 (t+1)^{1/2} ?

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

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integral of 9/2 (t+1)^{1/2}

\int \frac{9}{2} (t + 1)^{\frac{1}{2}} d t

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

Solution steps

\int \frac{9}{2} (t + 1)^{\frac{1}{2}} d t

Take the constant out:

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

= \frac{9}{2} \cdot \int (t + 1)^{\frac{1}{2}} d t

Apply Integration By Parts

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

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

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

Simplify

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

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

Add a constant to the solution

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

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