What is the integral from 0 to 24 of 6/(sqrt(x+1)) ?

The integral from 0 to 24 of 6/(sqrt(x+1)) is 48

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integral from 0 to 24 of 6/(sqrt(x+1))

\int_{0}^{24} \frac{6}{x + 1} d x

48

Solution steps

\int_{0}^{24} \frac{6}{x + 1} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{24} \frac{1}{x + 1} d x

Apply u-substitution

= 6 \cdot \int_{1}^{5} 2 d u

Integral of a constant:

\int a d x = a x

= 6 [2 u]_{1}^{5}

Compute the boundaries:

8

= 6 \cdot 8

Simplify

= 48

What is the integral from 0 to 24 of 6/(sqrt(x+1)) ?
The integral from 0 to 24 of 6/(sqrt(x+1)) is 48