integral of x/((x^2+1)sqrt(1-x^2))

Solution

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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