What is the integral of (sqrt(2x)+1)^3sqrt(2) ?

The integral of (sqrt(2x)+1)^3sqrt(2) is (sqrt(2))/5 (4sqrt(2)x^{5/2}+15x^2+10sqrt(2)x^{3/2}+5x)+C

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integral of (sqrt(2x)+1)^3sqrt(2)

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

\frac{2}{5} (42 x^{\frac{5}{2}} + 15 x^{2} + 102 x^{\frac{3}{2}} + 5 x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

= 2 \frac{1}{2} \cdot \int u^{\frac{3}{2}} + 3 u + 3 u + 1 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

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

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

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

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

\int 1 d u = u

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

Substitute back

u = 2 x

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

Simplify

2 \frac{1}{2} (\frac{2}{5} (2 x)^{\frac{5}{2}} + \frac{3 ( 2 x ) ^{2}}{2} + 2 (2 x)^{\frac{3}{2}} + 2 x) : \frac{2}{5} (42 x^{\frac{5}{2}} + 15 x^{2} + 102 x^{\frac{3}{2}} + 5 x)

= \frac{2}{5} (42 x^{\frac{5}{2}} + 15 x^{2} + 102 x^{\frac{3}{2}} + 5 x)

Add a constant to the solution

= \frac{2}{5} (42 x^{\frac{5}{2}} + 15 x^{2} + 102 x^{\frac{3}{2}} + 5 x) + C

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\int 40 x d x

What is the integral of (sqrt(2x)+1)^3sqrt(2) ?
The integral of (sqrt(2x)+1)^3sqrt(2) is (sqrt(2))/5 (4sqrt(2)x^{5/2}+15x^2+10sqrt(2)x^{3/2}+5x)+C