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

The integral from 1 to 4 of xsqrt(x^2+1) is (17sqrt(17)-2sqrt(2))/3

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

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

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

Decimal Notation

22.42145 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{17} u^{2} d u

= \int_{2}^{17} u^{2} d u

Apply the Power Rule

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

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

Compute the boundaries:

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

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

\int_{1}^{4} \frac{1}{6 x} d x

\int_{2}^{3} t d t

\int_{0}^{2} (8 - x^{3}) d x

\int_{π}^{2 π} 1 d x

\int_{1}^{2} x (x^{2} - 1)^{4} d x

What is the integral from 1 to 4 of xsqrt(x^2+1) ?
The integral from 1 to 4 of xsqrt(x^2+1) is (17sqrt(17)-2sqrt(2))/3