What is the integral from 0 to 1 of x^2sqrt(6x+8) ?

The integral from 0 to 1 of x^2sqrt(6x+8) is (2(833sqrt(14)-1024sqrt(2)))/(2835)

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integral from 0 to 1 of x^2sqrt(6x+8)

\int_{0}^{1} x^{2} 6 x + 8 d x

\frac{2 ( 833 14 - 1024 2 )}{2835}

Decimal Notation

1.17717 \dots

Solution steps

\int_{0}^{1} x^{2} 6 x + 8 d x

Apply u-substitution

: \int_{8}^{14} \frac{u ( u - 8 ) ^{2}}{216} d u

= \int_{8}^{14} \frac{u ( u - 8 ) ^{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_{8}^{14} u (u - 8)^{2} d u

Expand

u (u - 8)^{2} : u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 u

= \frac{1}{216} \cdot \int_{8}^{14} u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 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_{8}^{14} u^{\frac{5}{2}} d u - \int_{8}^{14} 16 u^{\frac{3}{2}} d u + \int_{8}^{14} 64 u d u)

\int_{8}^{14} u^{\frac{5}{2}} d u = 78414 - \frac{2048 2}{7}

\int_{8}^{14} 16 u^{\frac{3}{2}} d u = \frac{16 ( 392 14 - 256 2 )}{5}

\int_{8}^{14} 64 u d u = \frac{64 ( 28 14 - 32 2 )}{3}

= \frac{1}{216} (78414 - \frac{2048 2}{7} - \frac{16 ( 392 14 - 256 2 )}{5} + \frac{64 ( 28 14 - 32 2 )}{3})

Simplify

\frac{1}{216} (78414 - \frac{2048 2}{7} - \frac{16 ( 392 14 - 256 2 )}{5} + \frac{64 ( 28 14 - 32 2 )}{3}) : \frac{2 ( 833 14 - 1024 2 )}{2835}

= \frac{2 ( 833 14 - 1024 2 )}{2835}

\int \frac{( 7 x ^{2} + 5 x + 7 )}{( x ^{2} + 1 ) ^{2}} d x

\frac{d}{d x} (\frac{16}{x ^{2}} - \frac{4}{x})

f (x) = (a^{\frac{1}{x}} + b^{\frac{1}{x}})^{x}

\int \frac{x ^{3}}{5 x ^{2} + 2} d x

\frac{8}{s ^{2} + 9}

What is the integral from 0 to 1 of x^2sqrt(6x+8) ?
The integral from 0 to 1 of x^2sqrt(6x+8) is (2(833sqrt(14)-1024sqrt(2)))/(2835)