What is the integral from 0 to 4 of 1/(1+sqrt(t)) ?

The integral from 0 to 4 of 1/(1+sqrt(t)) is 2(2-ln(3))

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integral from 0 to 4 of 1/(1+sqrt(t))

\int_{0}^{4} \frac{1}{1 + t} d t

2 (2 - ln (3))

Decimal

1.80277 \dots

Solution steps

\int_{0}^{4} \frac{1}{1 + t} d t

Apply u-substitution

: \int_{1}^{3} \frac{2 ( u - 1 )}{u} d u

= \int_{1}^{3} \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_{1}^{3} \frac{u - 1}{u} d u

Expand

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

= 2 \cdot \int_{1}^{3} 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_{1}^{3} 1 d u - \int_{1}^{3} \frac{1}{u} d u)

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

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

= 2 (2 - ln (3))

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\int_{0}^{9} (9 x - x^{2}) d x

\int_{2}^{3} 8 x - 24 d x

What is the integral from 0 to 4 of 1/(1+sqrt(t)) ?
The integral from 0 to 4 of 1/(1+sqrt(t)) is 2(2-ln(3))