What is the integral of 6sqrt(x+1) ?

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

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integral of 6sqrt(x+1)

\int 6 x + 1 d x

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

Solution steps

\int 6 x + 1 d x

Take the constant out:

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

= 6 \cdot \int x + 1 d x

Apply u-substitution

: \int u d u

= 6 \cdot \int u 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 = x + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{e ^{t} - cos ( t )}{t}

\int - 2 x^{- 1} \cdot x^{- 2} d x

y^{^{''}} = - k^{2} \cdot y

f (x) = e^{\frac{x}{3}}

\int x e^{\frac{1}{2} x} d x

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