What is the integral of 3xsqrt(x+1) ?

The integral of 3xsqrt(x+1) is 6/5 (x+1)^{5/2}-2(x+1)^{3/2}+C

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integral of 3xsqrt(x+1)

\int 3 x x + 1 d x

\frac{6}{5} (x + 1)^{\frac{5}{2}} - 2 (x + 1)^{\frac{3}{2}} + C

Solution steps

\int 3 x x + 1 d x

Take the constant out:

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

= 3 \cdot \int x x + 1 d x

Apply u-substitution

: \int (u - 1) u d u

= 3 \cdot \int (u - 1) u d u

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= 3 (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = x + 1

= 3 (\frac{2}{5} (x + 1)^{\frac{5}{2}} - \frac{2}{3} (x + 1)^{\frac{3}{2}})

Simplify

3 (\frac{2}{5} (x + 1)^{\frac{5}{2}} - \frac{2}{3} (x + 1)^{\frac{3}{2}}) : \frac{6}{5} (x + 1)^{\frac{5}{2}} - 2 (x + 1)^{\frac{3}{2}}

= \frac{6}{5} (x + 1)^{\frac{5}{2}} - 2 (x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{6}{5} (x + 1)^{\frac{5}{2}} - 2 (x + 1)^{\frac{3}{2}} + C

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What is the integral of 3xsqrt(x+1) ?
The integral of 3xsqrt(x+1) is 6/5 (x+1)^{5/2}-2(x+1)^{3/2}+C