What is the integral from 1 to 8 of ((x^{-1/2})/(1+x^{1/2)}) ?

The integral from 1 to 8 of ((x^{-1/2})/(1+x^{1/2)}) is ln((9+4*2^{1/2})/4)

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integral from 1 to 8 of ((x^{-1/2})/(1+x^{1/2)})

\int_{1}^{8} (\frac{x ^{- \frac{1}{2}}}{1 + x ^{\frac{1}{2}}}) d x

ln (\frac{9 + 4 \cdot 2 ^{\frac{1}{2}}}{4})

Decimal

1.29861 \dots

Solution steps

\int_{1}^{8} \frac{x ^{- \frac{1}{2}}}{1 + x ^{\frac{1}{2}}} d x

Simplify

\frac{x ^{- \frac{1}{2}}}{1 + x ^{\frac{1}{2}}} : \frac{1}{x ^{\frac{1}{2}} + x}

= \int_{1}^{8} \frac{1}{x ^{\frac{1}{2}} + x} d x

Apply u-substitution

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

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

Apply u-substitution

: \int_{2}^{1 + 2 \cdot 2^{\frac{1}{2}}} \frac{1}{v} d v

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 [ln ∣ v ∣]_{2}^{1 + 2 \cdot 2^{\frac{1}{2}}}

Compute the boundaries:

ln (1 + 2 \cdot 2^{\frac{1}{2}}) - ln (2)

= 2 (ln (1 + 2 \cdot 2^{\frac{1}{2}}) - ln (2))

Simplify

= ln (\frac{9 + 4 \cdot 2 ^{\frac{1}{2}}}{4})

\int \frac{5 r ^{2}}{r + 5} d r

\int e^{x - y} d x

\int t sec^{2} (8 t) d t

\frac{d ^{2} y}{d t ^{2}} + 7 \frac{d y}{d t} + 12 y = 0

\int_{2}^{5} e^{1.5 t} - 2 t d t

What is the integral from 1 to 8 of ((x^{-1/2})/(1+x^{1/2)}) ?
The integral from 1 to 8 of ((x^{-1/2})/(1+x^{1/2)}) is ln((9+4*2^{1/2})/4)