What is the integral of 1/(1-t^{1/2)} ?

The integral of 1/(1-t^{1/2)} is -2(-1+t^{1/2}+ln|1-t^{1/2}|)+C

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integral of 1/(1-t^{1/2)}

\int \frac{1}{1 - t ^{\frac{1}{2}}} d t

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

Solution steps

\int \frac{1}{1 - t ^{\frac{1}{2}}} d t

Apply u-substitution

: \int \frac{2 u}{1 - u} d u

= \int \frac{2 u}{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 \frac{u}{1 - u} d u

Apply u-substitution

: \int - \frac{- v + 1}{v} d v

= 2 \cdot \int - \frac{- v + 1}{v} d v

Take the constant out:

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

= 2 (- \int \frac{- v + 1}{v} d v)

Expand

\frac{- v + 1}{v} : - 1 + \frac{1}{v}

= 2 (- \int - 1 + \frac{1}{v} d v)

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 d v + \int \frac{1}{v} d v))

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (- (- v + ln ∣ v ∣))

Substitute back

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

Simplify

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

Add a constant to the solution

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

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