What is the integral from 5 to 1 of sqrt(2x)ln(x^3) ?

The integral from 5 to 1 of sqrt(2x)ln(x^3) is -sqrt(2)(10sqrt(5)ln(5)+(-20sqrt(5)+4)/3)

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integral from 5 to 1 of sqrt(2x)ln(x^3)

\int_{5}^{1} 2 x ln (x^{3}) d x

- 2 (105 ln (5) + \frac{- 20 5 + 4}{3})

Decimal Notation

- 31.69866 \dots

Solution steps

\int_{5}^{1} 2 x ln (x^{3}) d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{5} 2 x ln (x^{3}) d x

2 x = 2 x

= - \int_{1}^{5} 2 x ln (x^{3}) d x

Take the constant out:

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

= - 2 \cdot \int_{1}^{5} x ln (x^{3}) d x

Apply Integration By Parts

: [\frac{2}{3} x^{\frac{3}{2}} ln (x^{3}) - \int 2 x^{\frac{1}{2}} d x]_{1}^{5}

= - 2 [\frac{2}{3} x^{\frac{3}{2}} ln (x^{3}) - \int 2 x^{\frac{1}{2}} d x]_{1}^{5}

\int 2 x^{\frac{1}{2}} d x = \frac{4}{3} x^{\frac{3}{2}}

= - 2 [\frac{2}{3} x^{\frac{3}{2}} ln (x^{3}) - \frac{4}{3} x^{\frac{3}{2}}]_{1}^{5}

\frac{2}{3} x^{\frac{3}{2}} ln (x^{3}) = 2 x^{\frac{3}{2}} ln (x)

= - 2 [2 x^{\frac{3}{2}} ln (x) - \frac{4}{3} x^{\frac{3}{2}}]_{1}^{5}

Compute the boundaries:

105 ln (5) - \frac{20 5}{3} + \frac{4}{3}

= - 2 (105 ln (5) - \frac{20 5}{3} + \frac{4}{3})

Simplify

= - 2 (105 ln (5) + \frac{- 20 5 + 4}{3})

\int_{0}^{2} - x^{2} + 2 x d x

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\int_{- 1}^{1} x + y + 1 d x

\int_{1}^{x^{3}} e^{t^{2}} d t

What is the integral from 5 to 1 of sqrt(2x)ln(x^3) ?
The integral from 5 to 1 of sqrt(2x)ln(x^3) is -sqrt(2)(10sqrt(5)ln(5)+(-20sqrt(5)+4)/3)