What is the integral from 1 to 16 of 1/(10+sqrt(x)) ?

The integral from 1 to 16 of 1/(10+sqrt(x)) is 2(3-10(ln(14)-ln(11)))

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integral from 1 to 16 of 1/(10+sqrt(x))

\int_{1}^{16} \frac{1}{10 + x} d x

2 (3 - 10 (ln (14) - ln (11)))

Decimal Notation

1.17675 \dots

Solution steps

\int_{1}^{16} \frac{1}{10 + x} d x

Apply u-substitution

: \int_{11}^{14} \frac{2 ( u - 10 )}{u} d u

= \int_{11}^{14} \frac{2 ( u - 10 )}{u} d u

Take the constant out:

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

= 2 \cdot \int_{11}^{14} \frac{u - 10}{u} d u

Expand

\frac{u - 10}{u} : 1 - \frac{10}{u}

= 2 \cdot \int_{11}^{14} 1 - \frac{10}{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_{11}^{14} 1 d u - \int_{11}^{14} \frac{10}{u} d u)

\int_{11}^{14} 1 d u = 3

\int_{11}^{14} \frac{10}{u} d u = 10 (ln (14) - ln (11))

= 2 (3 - 10 (ln (14) - ln (11)))

x \to 1 + lim (\frac{3 + x}{x ^{2} - 1})

f (x) = \frac{4 x}{x ^{2} + 1}, at x = - 1

\int \frac{x}{x ^{2} + 3 x + 3} d x

8 tan^{2} (8 x)

What is the integral from 1 to 16 of 1/(10+sqrt(x)) ?
The integral from 1 to 16 of 1/(10+sqrt(x)) is 2(3-10(ln(14)-ln(11)))