What is the integral from 1 to 2 of 1/(sqrt(t)+e) ?

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

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integral from 1 to 2 of 1/(sqrt(t)+e)

\int_{1}^{2} \frac{1}{t + e} d t

2 (2 - 1 - e (ln (2 + e) - ln (1 + e)))

Decimal

0.25421 \dots

Solution steps

\int_{1}^{2} \frac{1}{t + e} d t

Apply u-substitution

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

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

Expand

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

= 2 \cdot \int_{1 + e}^{2 + e} 1 - \frac{e}{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 + e}^{2 + e} 1 d u - \int_{1 + e}^{2 + e} \frac{e}{u} d u)

\int_{1 + e}^{2 + e} 1 d u = 2 - 1

\int_{1 + e}^{2 + e} \frac{e}{u} d u = e (ln (2 + e) - ln (1 + e))

= 2 (2 - 1 - e (ln (2 + e) - ln (1 + e)))

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