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

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

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

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

2 (2 - ln (3))

Decimal

1.80277 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{2} \frac{2 u}{1 + u} d u

= \int_{0}^{2} \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_{0}^{2} \frac{u}{1 + u} d u

Apply u-substitution

: \int_{1}^{3} \frac{v - 1}{v} d v

= 2 \cdot \int_{1}^{3} \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= 2 \cdot \int_{1}^{3} 1 - \frac{1}{v} d v

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_{1}^{3} 1 d v - \int_{1}^{3} \frac{1}{v} d v)

\int_{1}^{3} 1 d v = 2

\int_{1}^{3} \frac{1}{v} d v = ln (3)

= 2 (2 - ln (3))

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

\int_{0}^{l n (2)} (e^{t} - e^{- t}) d t

\int_{1}^{6} 7 x + 1 d x

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

\int_{- 2}^{2} y^{2} d y

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