What is the integral from 0 to 2 of x^3sqrt(3x+1) ?

The integral from 0 to 2 of x^3sqrt(3x+1) is (92512sqrt(7)+32)/(25515)

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integral from 0 to 2 of x^3sqrt(3x+1)

\int_{0}^{2} x^{3} 3 x + 1 d x

\frac{92512 7 + 32}{25515}

Decimal Notation

9.59418 \dots

Solution steps

\int_{0}^{2} x^{3} 3 x + 1 d x

Apply u-substitution

: \int_{1}^{7} \frac{u ( u - 1 ) ^{3}}{81} d u

= \int_{1}^{7} \frac{u ( u - 1 ) ^{3}}{81} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int_{1}^{7} u (u - 1)^{3} d u

Expand

u (u - 1)^{3} : u^{\frac{7}{2}} - 3 u^{\frac{5}{2}} + 3 u^{\frac{3}{2}} - u

= \frac{1}{81} \cdot \int_{1}^{7} u^{\frac{7}{2}} - 3 u^{\frac{5}{2}} + 3 u^{\frac{3}{2}} - 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}{81} (\int_{1}^{7} u^{\frac{7}{2}} d u - \int_{1}^{7} 3 u^{\frac{5}{2}} d u + \int_{1}^{7} 3 u^{\frac{3}{2}} d u - \int_{1}^{7} u d u)

\int_{1}^{7} u^{\frac{7}{2}} d u = \frac{4802 7 - 2}{9}

\int_{1}^{7} 3 u^{\frac{5}{2}} d u = 3 (987 - \frac{2}{7})

\int_{1}^{7} 3 u^{\frac{3}{2}} d u = \frac{3 ( 98 7 - 2 )}{5}

\int_{1}^{7} u d u = \frac{14 7 - 2}{3}

= \frac{1}{81} (\frac{4802 7 - 2}{9} - 3 (987 - \frac{2}{7}) + \frac{3 ( 98 7 - 2 )}{5} - \frac{14 7 - 2}{3})

Simplify

\frac{1}{81} (\frac{4802 7 - 2}{9} - 3 (987 - \frac{2}{7}) + \frac{3 ( 98 7 - 2 )}{5} - \frac{14 7 - 2}{3}) : \frac{92512 7 + 32}{25515}

= \frac{92512 7 + 32}{25515}

ln (x), at 4

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

n = 1 \sum \infty \frac{( e ^{n} )}{n}

y^{^{'''}} + 13 y^{^{''}} - 9 y^{^{'}} - 405 y = 0

y^{^{'}} = 3 - \frac{2 y}{200 + x}

What is the integral from 0 to 2 of x^3sqrt(3x+1) ?
The integral from 0 to 2 of x^3sqrt(3x+1) is (92512sqrt(7)+32)/(25515)