What is the integral from 0 to 5 of 1/(sqrt(3x+1)) ?

The integral from 0 to 5 of 1/(sqrt(3x+1)) is 2

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integral from 0 to 5 of 1/(sqrt(3x+1))

\int_{0}^{5} \frac{1}{3 x + 1} d x

2

Solution steps

\int_{0}^{5} \frac{1}{3 x + 1} d x

Apply u-substitution

: \int_{1}^{16} \frac{1}{3 u} d u

= \int_{1}^{16} \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_{1}^{16} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{1}^{16}

= \frac{1}{3} [2 u^{\frac{1}{2}}]_{1}^{16}

Apply radical rule:

= \frac{1}{3} [2 u]_{1}^{16}

Compute the boundaries:

6

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

Simplify

= 2

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

\int 9 (2 x - 9)^{8} d x

(1 + x)^{n}

\frac{d}{d x} (x^{2^{x}})

y^{^{''}} - (\frac{2}{t}) y^{^{'}} + (\frac{2}{t ^{2}}) y = \frac{2 t}{1 + t ^{2}}

What is the integral from 0 to 5 of 1/(sqrt(3x+1)) ?
The integral from 0 to 5 of 1/(sqrt(3x+1)) is 2