What is the integral from 0 to 4 of x/(sqrt(5x^2+1)) ?

The integral from 0 to 4 of x/(sqrt(5x^2+1)) is 8/5

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integral from 0 to 4 of x/(sqrt(5x^2+1))

\int_{0}^{4} \frac{x}{5 x ^{2} + 1} d x

\frac{8}{5}

Decimal

1.6

Solution steps

\int_{0}^{4} \frac{x}{5 x ^{2} + 1} d x

Apply u-substitution

: \int_{1}^{81} \frac{1}{10 u} d u

= \int_{1}^{81} \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_{1}^{81} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{81}

= \frac{1}{10} [2 u^{\frac{1}{2}}]_{1}^{81}

Apply radical rule:

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

= \frac{1}{10} [2 u]_{1}^{81}

Compute the boundaries:

16

= \frac{1}{10} \cdot 16

Simplify

= \frac{8}{5}

\int_{1}^{2} (1 + ln (x)) x^{x} d x

\int_{- 1}^{5} \frac{x}{5 + x} d x

\int_{2}^{\infty} ln (x) d x

\int_{e}^{e^{2}} \frac{1}{x ln ^{3} ( x )} d x

\int_{1}^{2} \frac{1}{9} x^{2} d x

What is the integral from 0 to 4 of x/(sqrt(5x^2+1)) ?
The integral from 0 to 4 of x/(sqrt(5x^2+1)) is 8/5