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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

(u + 1)^{2} : u + 2 u + 1

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

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

\int 2 u d u = \frac{4}{3} u^{\frac{3}{2}}

\int 1 d u = u

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

Substitute back

u = 2 x

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

Simplify

2 \frac{1}{2} (\frac{( 2 x ) ^{2}}{2} + \frac{4}{3} (2 x)^{\frac{3}{2}} + 2 x) : \frac{2}{3} (3 x^{2} + 42 x^{\frac{3}{2}} + 3 x)

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

Add a constant to the solution

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

\int (\frac{ln ( x )}{x} + sec^{2} (2 x)) d x

\int x (x) d x

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

\int \frac{x ^{2}}{1 - 3 x ^{2}} d x

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