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

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

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integral from 1 to 8 of 1/(sqrt(3x+1))

\int_{1}^{8} \frac{1}{3 x + 1} d x

2

Solution steps

\int_{1}^{8} \frac{1}{3 x + 1} d x

Apply u-substitution

: \int_{4}^{25} \frac{1}{3 u} d u

= \int_{4}^{25} \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{4}^{25} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{4}^{25}

= \frac{1}{3} [2 u^{\frac{1}{2}}]_{4}^{25}

Apply radical rule:

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

= \frac{1}{3} [2 u]_{4}^{25}

Compute the boundaries:

6

= \frac{1}{3} \cdot 6

Simplify

= 2

x \to - 1 lim (\frac{1}{x + 1})

y^{^{''}} - y^{^{'}} + 4 y = 2 sin (2 t)

3 x \cdot ln (\frac{1}{x ^{2}})

\int_{4}^{\infty} x e^{- 5 x} d x

(y^{2} + x y^{2}) \frac{d y}{d x} = 1

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