What is the integral from 1 to 9 of 1/(1+sqrt(x)) ?

The integral from 1 to 9 of 1/(1+sqrt(x)) is 2(2-ln(2))

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integral from 1 to 9 of 1/(1+sqrt(x))

\int_{1}^{9} \frac{1}{1 + x} d x

2 (2 - ln (2))

Decimal Notation

2.61370 \dots

Solution steps

\int_{1}^{9} \frac{1}{1 + x} d x

Apply u-substitution

: \int_{2}^{4} \frac{2 ( u - 1 )}{u} d u

= \int_{2}^{4} \frac{2 ( u - 1 )}{u} d u

Take the constant out:

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

= 2 \cdot \int_{2}^{4} \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 2 \cdot \int_{2}^{4} 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int_{2}^{4} 1 d u - \int_{2}^{4} \frac{1}{u} d u)

\int_{2}^{4} 1 d u = 2

\int_{2}^{4} \frac{1}{u} d u = ln (2)

= 2 (2 - ln (2))

\int_{2}^{4} ln (x) d x

\int_{0}^{\frac{π}{2}} cos^{5} (t) d t

\int_{0}^{4} 5 x d x

\int_{0}^{\infty} sin (\frac{x}{2}) d x

\int_{0}^{π} sin (x) + cos (x) d x

What is the integral from 1 to 9 of 1/(1+sqrt(x)) ?
The integral from 1 to 9 of 1/(1+sqrt(x)) is 2(2-ln(2))