What is the integral of x^2sqrt(6x+7) ?

The integral of x^2sqrt(6x+7) is 1/2835 (135x^2-126x+98)(6x+7)^{3/2}+C

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integral of x^2sqrt(6x+7)

\int x^{2} 6 x + 7 d x

\frac{1}{2835} (135 x^{2} - 126 x + 98) (6 x + 7)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 6 x + 7 d x

Apply u-substitution

: \int \frac{u ( u - 7 ) ^{2}}{216} d u

= \int \frac{u ( u - 7 ) ^{2}}{216} d u

Take the constant out:

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

= \frac{1}{216} \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}{216} \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}{216} (\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}{216} (\frac{2}{7} u^{\frac{7}{2}} - \frac{28}{5} u^{\frac{5}{2}} + \frac{98}{3} u^{\frac{3}{2}})

Substitute back

u = 6 x + 7

= \frac{1}{216} (\frac{2}{7} (6 x + 7)^{\frac{7}{2}} - \frac{28}{5} (6 x + 7)^{\frac{5}{2}} + \frac{98}{3} (6 x + 7)^{\frac{3}{2}})

Simplify

\frac{1}{216} (\frac{2}{7} (6 x + 7)^{\frac{7}{2}} - \frac{28}{5} (6 x + 7)^{\frac{5}{2}} + \frac{98}{3} (6 x + 7)^{\frac{3}{2}}) : \frac{1}{2835} (135 x^{2} - 126 x + 98) (6 x + 7)^{\frac{3}{2}}

= \frac{1}{2835} (135 x^{2} - 126 x + 98) (6 x + 7)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{2835} (135 x^{2} - 126 x + 98) (6 x + 7)^{\frac{3}{2}} + C

f (x) = e

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

\frac{d}{d x} (tan (e^{3 x}) + e^{t a n (3 x)})

\frac{\partial}{\partial x} (250 - 4 x - 7 y + 0.2 x^{2} + 0.1 y^{2})

\int \frac{3}{2} x + 1 d x

What is the integral of x^2sqrt(6x+7) ?
The integral of x^2sqrt(6x+7) is 1/2835 (135x^2-126x+98)(6x+7)^{3/2}+C