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

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

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integral of 3(t+1)^{1/2}

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

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

Solution steps

\int 3 (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

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

Apply Integration By Parts

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

= 3 (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}}

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

Simplify

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

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

Add a constant to the solution

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

x \to 0.5 lim (3 x^{2} + 2 x - 3)

\frac{\partial}{\partial x} (x^{\frac{1}{3}} y^{\frac{2}{3}})

x^{2}, - x^{2} + 4 x

\int \frac{1}{2 - 5 x} d x

x \to 0 lim (x (x - 1))

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