What is the integral of 1/((x^{-0)0.5+x^{0.5})} ?

The integral of 1/((x^{-0)0.5+x^{0.5})} is 2(-0.5ln|0.5+sqrt(x)|+sqrt(x))+C

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integral of 1/((x^{-0)0.5+x^{0.5})}

integral

\frac{1}{( x ^{- 0} 0.5 + x ^{0.5} )}

2 (- 0.5 ln 0.5 + x + x) + C

Solution steps

\int \frac{1}{x ^{- 0} \cdot 0.5 + x ^{0.5}} d x

Apply u-substitution

: \int \frac{u}{0.5 ( 0.5 + u )} d u

= \int \frac{u}{0.5 ( 0.5 + u )} d u

Take the constant out:

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

= \frac{1}{0.5} \cdot \int \frac{u}{0.5 + u} d u

Simplify

= 2 \cdot \int \frac{u}{0.5 + u} d u

\frac{u}{0.5 + u} = - \frac{0.5}{0.5 + u} + 1

= 2 \cdot \int - \frac{0.5}{0.5 + u} + 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 \frac{0.5}{0.5 + u} d u + \int 1 d u)

\int \frac{0.5}{0.5 + u} d u = 0.5 ln ∣ 0.5 + u ∣

\int 1 d u = u

= 2 (- 0.5 ln ∣ 0.5 + u ∣ + u)

Substitute back

u = x^{0.5}

= 2 (- 0.5 ln 0.5 + x^{0.5} + x^{0.5})

Apply radical rule:

= 2 (- 0.5 ln 0.5 + x + x)

Add a constant to the solution

= 2 (- 0.5 ln 0.5 + x + x) + C

\frac{e ^{1}}{e ^{x} + 1}

\frac{ln ( 1 + x ^{2} )}{x ^{2}}

\frac{1}{t ^{2}}

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

2^{2 x}

What is the integral of 1/((x^{-0)0.5+x^{0.5})} ?
The integral of 1/((x^{-0)0.5+x^{0.5})} is 2(-0.5ln|0.5+sqrt(x)|+sqrt(x))+C