What is the integral from 0 to 3 of x(8x^2+9)^{1/2} ?

The integral from 0 to 3 of x(8x^2+9)^{1/2} is 117/4

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integral from 0 to 3 of x(8x^2+9)^{1/2}

\int_{0}^{3} x (8 x^{2} + 9)^{\frac{1}{2}} d x

\frac{117}{4}

Decimal

29.25

Solution steps

\int_{0}^{3} x (8 x^{2} + 9)^{\frac{1}{2}} d x

Multiply by the conjugate of

8 x^{2} + 9 : \frac{64 x ^{4} - 81}{8 x ^{2} - 9}

= \int_{0}^{3} x (\frac{64 x ^{4} - 81}{8 x ^{2} - 9})^{\frac{1}{2}} d x

Find undefined (singularity) points:

x = \frac{3}{2 2}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{\frac{3}{2 2}} x (\frac{64 x ^{4} - 81}{8 x ^{2} - 9})^{\frac{1}{2}} d x + \int_{\frac{3}{2 2}}^{3} x (\frac{64 x ^{4} - 81}{8 x ^{2} - 9})^{\frac{1}{2}} d x

\int_{0}^{\frac{3}{2 2}} x (\frac{64 x ^{4} - 81}{8 x ^{2} - 9})^{\frac{1}{2}} d x = \frac{9 ( 2 2 - 1 )}{8}

\int_{\frac{3}{2 2}}^{3} x (\frac{64 x ^{4} - 81}{8 x ^{2} - 9})^{\frac{1}{2}} d x = \frac{9 ( 27 - 2 2 )}{8}

= \frac{9 ( 2 2 - 1 )}{8} + \frac{9 ( 27 - 2 2 )}{8}

Simplify

\frac{9 ( 2 2 - 1 )}{8} + \frac{9 ( 27 - 2 2 )}{8} : \frac{117}{4}

= \frac{117}{4}

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What is the integral from 0 to 3 of x(8x^2+9)^{1/2} ?
The integral from 0 to 3 of x(8x^2+9)^{1/2} is 117/4