What is the integral from 2 to 6 of x/(sqrt(5x^2+1)) ?

The integral from 2 to 6 of x/(sqrt(5x^2+1)) is (sqrt(181)-sqrt(21))/5

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integral from 2 to 6 of x/(sqrt(5x^2+1))

\int_{2}^{6} \frac{x}{5 x ^{2} + 1} d x

\frac{181 - 21}{5}

Decimal

1.77420 \dots

Solution steps

\int_{2}^{6} \frac{x}{5 x ^{2} + 1} d x

Apply u-substitution

: \int_{21}^{181} \frac{1}{10 u} d u

= \int_{21}^{181} \frac{1}{10 u} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int_{21}^{181} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{21}^{181}

= \frac{1}{10} [2 u^{\frac{1}{2}}]_{21}^{181}

Apply radical rule:

a^{\frac{1}{n}} = n a

= \frac{1}{10} [2 u]_{21}^{181}

Compute the boundaries:

2181 - 221

= \frac{1}{10} (2181 - 221)

Simplify

= \frac{181 - 21}{5}

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What is the integral from 2 to 6 of x/(sqrt(5x^2+1)) ?
The integral from 2 to 6 of x/(sqrt(5x^2+1)) is (sqrt(181)-sqrt(21))/5