What is the integral of x^2sqrt(4x+7) ?

The integral of x^2sqrt(4x+7) is 1/420 (30x^2-42x+49)(4x+7)^{3/2}+C

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integral of x^2sqrt(4x+7)

\int x^{2} 4 x + 7 d x

\frac{1}{420} (30 x^{2} - 42 x + 49) (4 x + 7)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 4 x + 7 d x

Apply u-substitution

: \int \frac{u ( u - 7 ) ^{2}}{64} d u

= \int \frac{u ( u - 7 ) ^{2}}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int u (u - 7)^{2} d u

Expand

u (u - 7)^{2} : u^{\frac{5}{2}} - 14 u^{\frac{3}{2}} + 49 u

= \frac{1}{64} \cdot \int u^{\frac{5}{2}} - 14 u^{\frac{3}{2}} + 49 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

= \frac{1}{64} (\int u^{\frac{5}{2}} d u - \int 14 u^{\frac{3}{2}} d u + \int 49 u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 14 u^{\frac{3}{2}} d u = \frac{28}{5} u^{\frac{5}{2}}

\int 49 u d u = \frac{98}{3} u^{\frac{3}{2}}

= \frac{1}{64} (\frac{2}{7} u^{\frac{7}{2}} - \frac{28}{5} u^{\frac{5}{2}} + \frac{98}{3} u^{\frac{3}{2}})

Substitute back

u = 4 x + 7

= \frac{1}{64} (\frac{2}{7} (4 x + 7)^{\frac{7}{2}} - \frac{28}{5} (4 x + 7)^{\frac{5}{2}} + \frac{98}{3} (4 x + 7)^{\frac{3}{2}})

Simplify

\frac{1}{64} (\frac{2}{7} (4 x + 7)^{\frac{7}{2}} - \frac{28}{5} (4 x + 7)^{\frac{5}{2}} + \frac{98}{3} (4 x + 7)^{\frac{3}{2}}) : \frac{1}{420} (30 x^{2} - 42 x + 49) (4 x + 7)^{\frac{3}{2}}

= \frac{1}{420} (30 x^{2} - 42 x + 49) (4 x + 7)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{420} (30 x^{2} - 42 x + 49) (4 x + 7)^{\frac{3}{2}} + C

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What is the integral of x^2sqrt(4x+7) ?
The integral of x^2sqrt(4x+7) is 1/420 (30x^2-42x+49)(4x+7)^{3/2}+C