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

The integral of 1/(t-sqrt(1-t^2)) is

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integral of 1/(t-sqrt(1-t^2))

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{cos ( u )}{sin ( u ) - cos ( u )} d u

= \int \frac{cos ( u )}{sin ( u ) - cos ( u )} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1 - v ^{2}}{( 2 v - 1 + v ^{2} ) ( 1 + v ^{2} )} : \frac{v + 1}{2 ( v ^{2} + 2 v - 1 )} + \frac{- v - 1}{2 ( v ^{2} + 1 )}

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

\int \frac{v + 1}{2 ( v ^{2} + 2 v - 1 )} d v = \frac{1}{4} ln v^{2} + 2 v - 1

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

= 2 (\frac{1}{4} ln v^{2} + 2 v - 1 + \frac{1}{2} (- \frac{1}{2} ln v^{2} + 1 - arctan (v)))

Substitute back

= 2 (\frac{1}{4} ln tan^{2} (\frac{arcsin ( t )}{2}) + 2 tan (\frac{arcsin ( t )}{2}) - 1 + \frac{1}{2} (- \frac{1}{2} ln tan^{2} (\frac{arcsin ( t )}{2}) + 1 - arctan (tan (\frac{arcsin ( t )}{2}))))

Simplify

2 (\frac{1}{4} ln tan^{2} (\frac{arcsin ( t )}{2}) + 2 tan (\frac{arcsin ( t )}{2}) - 1 + \frac{1}{2} (- \frac{1}{2} ln tan^{2} (\frac{arcsin ( t )}{2}) + 1 - arctan (tan (\frac{arcsin ( t )}{2})))) : \frac{1}{2} ln tan^{2} (\frac{1}{2} arcsin (t)) + 2 tan (\frac{1}{2} arcsin (t)) - 1 - \frac{1}{2} ln tan^{2} (\frac{1}{2} arcsin (t)) + 1 - arctan (tan (\frac{1}{2} arcsin (t)))

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

Add a constant to the solution

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

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What is the integral of 1/(t-sqrt(1-t^2)) ?
The integral of 1/(t-sqrt(1-t^2)) is