What is the integral from 3 to 8 of (sqrt(x+1))/x ?

The integral from 3 to 8 of (sqrt(x+1))/x is -ln(2)+ln(3)+2

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integral from 3 to 8 of (sqrt(x+1))/x

\int_{3}^{8} \frac{x + 1}{x} d x

- ln (2) + ln (3) + 2

Decimal Notation

2.40546 \dots

Solution steps

\int_{3}^{8} \frac{x + 1}{x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Take the constant out:

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

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

\frac{u ^{2}}{- u ^{2} + 1} = \frac{1}{- u ^{2} + 1} - 1

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

\int_{2}^{3} \frac{1}{- u ^{2} + 1} d u = \frac{ln ( 2 ) - ln ( 3 )}{2}

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

= 2 (- (\frac{ln ( 2 ) - ln ( 3 )}{2} - 1))

Simplify

2 (- (\frac{ln ( 2 ) - ln ( 3 )}{2} - 1)) : - ln (2) + ln (3) + 2

= - ln (2) + ln (3) + 2

y = 3 x^{2}, y = 6 - 3 x

y = x, y = 5 x, y = - x + 4

\int \frac{2 x + 1}{x ^{2} + 2 x + 17} d x

\int_{0}^{1} 13 cos (x^{2}) d x

- csc (x)

What is the integral from 3 to 8 of (sqrt(x+1))/x ?
The integral from 3 to 8 of (sqrt(x+1))/x is -ln(2)+ln(3)+2