What is the integral from 0 to 5 of 1/(sqrt(7x+3)) ?

The integral from 0 to 5 of 1/(sqrt(7x+3)) is (2sqrt(38)-2sqrt(3))/7

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integral from 0 to 5 of 1/(sqrt(7x+3))

\int_{0}^{5} \frac{1}{7 x + 3} d x

\frac{2 38 - 2 3}{7}

Solution steps

\int_{0}^{5} \frac{1}{7 x + 3} d x

Apply u-substitution

: \int_{3}^{38} \frac{1}{7 u} d u

= \int_{3}^{38} \frac{1}{7 u} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int_{3}^{38} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{3}^{38}

= \frac{1}{7} [2 u^{\frac{1}{2}}]_{3}^{38}

Apply radical rule:

= \frac{1}{7} [2 u]_{3}^{38}

Compute the boundaries:

238 - 23

= \frac{1}{7} (238 - 23)

Simplify

= \frac{2 38 - 2 3}{7}

\int_{- \infty}^{0} \frac{x}{x ^{4} + 1} d x

\int_{- 1}^{0} \frac{1}{1 - x} d x

\int_{1}^{6} 4 x^{3} - 4 x^{4} - 3 d x

\int_{0}^{2} 1 - \frac{1}{2} x d x

\int_{0}^{1} (2 x - x^{2}) - x^{2} d x

What is the integral from 0 to 5 of 1/(sqrt(7x+3)) ?
The integral from 0 to 5 of 1/(sqrt(7x+3)) is (2sqrt(38)-2sqrt(3))/7