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

The integral from 0 to 6 of 2xsqrt(x^2+1) is (2(37sqrt(37)-1))/3

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

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

\frac{2 ( 37 37 - 1 )}{3}

Decimal Notation

149.37480 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{1}^{37} u^{2} d u

= 2 \cdot \int_{1}^{37} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{1}^{37}

= 2 [\frac{u ^{3}}{3}]_{1}^{37}

Compute the boundaries:

\frac{37 37}{3} - \frac{1}{3}

= 2 (\frac{37 37}{3} - \frac{1}{3})

Simplify

= \frac{2 ( 37 37 - 1 )}{3}

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What is the integral from 0 to 6 of 2xsqrt(x^2+1) ?
The integral from 0 to 6 of 2xsqrt(x^2+1) is (2(37sqrt(37)-1))/3