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

The integral from 1 to 5 of 1/(sqrt(3x+1)) is 4/3

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

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{16} \frac{1}{3 u} d u

= \int_{4}^{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_{4}^{16} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{4}^{16}

= \frac{1}{3} [2 u^{\frac{1}{2}}]_{4}^{16}

Apply radical rule:

= \frac{1}{3} [2 u]_{4}^{16}

Compute the boundaries:

4

= \frac{1}{3} \cdot 4

Simplify

= \frac{4}{3}

\frac{d}{d x} (\frac{4 x ^{2} + 2 x + 2}{x})

\int 1 + tan^{2} (x) d x

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

6 y^{^{''}} + y^{^{'}} = 0

\frac{d}{d x} (\frac{π}{2})

What is the integral from 1 to 5 of 1/(sqrt(3x+1)) ?
The integral from 1 to 5 of 1/(sqrt(3x+1)) is 4/3