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

The integral from 1 to 8 of x/(sqrt(8+x^2)) is 3(2sqrt(2)-1)

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integral from 1 to 8 of x/(sqrt(8+x^2))

\int_{1}^{8} \frac{x}{8 + x ^{2}} d x

3 (22 - 1)

Decimal

5.48528 \dots

Solution steps

\int_{1}^{8} \frac{x}{8 + x ^{2}} d x

Apply u-substitution

: \int_{9}^{72} \frac{1}{2 u} d u

= \int_{9}^{72} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{9}^{72} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{9}^{72}

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{9}^{72}

Apply radical rule:

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

= \frac{1}{2} [2 u]_{9}^{72}

Compute the boundaries:

122 - 6

= \frac{1}{2} (122 - 6)

Simplify

= 3 (22 - 1)

x \to - 1 - lim (∣ x ∣ - 3)

\int \frac{e ^{8 x}}{e ^{5 x}} d x

(0, - 6), (5, - 3)

\frac{d}{d x} (x cos (x y))

\int_{6}^{8} \frac{1}{( x - 6 ) ^{3}} d x

What is the integral from 1 to 8 of x/(sqrt(8+x^2)) ?
The integral from 1 to 8 of x/(sqrt(8+x^2)) is 3(2sqrt(2)-1)