What is the integral of 6(t+1)^{1/2} ?

The integral of 6(t+1)^{1/2} is 4(t+1)^{3/2}+C

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integral of 6(t+1)^{1/2}

\int 6 (t + 1)^{\frac{1}{2}} d t

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

Solution steps

\int 6 (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

= 6 \cdot \int (t + 1)^{\frac{1}{2}} d t

Apply u-substitution

: \int u^{\frac{1}{2}} d u

= 6 \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

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

= 6 \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = t + 1

= 6 \cdot \frac{2}{3} (t + 1)^{\frac{3}{2}}

Simplify

6 \cdot \frac{2}{3} (t + 1)^{\frac{3}{2}} : 4 (t + 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

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What is the integral of 6(t+1)^{1/2} ?
The integral of 6(t+1)^{1/2} is 4(t+1)^{3/2}+C