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

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

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

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

4

Solution steps

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

Apply u-substitution

: \int_{0}^{2} 2 d u

= \int_{0}^{2} 2 d u

Integral of a constant:

\int a d x = a x

= [2 u]_{0}^{2}

Compute the boundaries:

4

= 4

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

\int_{0}^{s i n (y)} cos (y) d y

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

\int_{2}^{x} 5 t + 3 d t

\int_{0}^{1} x^{2} 1 + x d x

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