What is the integral from 0 to 2 of xsqrt(1+4x^2) ?

The integral from 0 to 2 of xsqrt(1+4x^2) is (17sqrt(17)-1)/(12)

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integral from 0 to 2 of xsqrt(1+4x^2)

\int_{0}^{2} x 1 + 4 x^{2} d x

\frac{17 17 - 1}{12}

Solution steps

\int_{0}^{2} x 1 + 4 x^{2} d x

Apply u-substitution

: \int_{1}^{17} \frac{u}{8} d u

= \int_{1}^{17} \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{17} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Compute the boundaries:

\frac{34 17}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{34 17}{3} - \frac{2}{3})

Simplify

= \frac{17 17 - 1}{12}

\int_{2}^{\infty} \frac{5}{4 + x ^{2}} d x

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\int_{0}^{3} 6 x - x^{2} d x

\int_{1}^{2} x^{2} + \frac{1}{x ^{4}} d x

What is the integral from 0 to 2 of xsqrt(1+4x^2) ?
The integral from 0 to 2 of xsqrt(1+4x^2) is (17sqrt(17)-1)/(12)