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

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

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

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

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

Solution steps

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

Simplify

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

= \int \frac{1}{( - x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int sec^{2} (u) d u

= \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= tan (u)

Substitute back

u = arcsin (x)

= tan (arcsin (x))

Use the following identity:

tan (arcsin (x)) = \frac{x}{1 - x ^{2}}

= \frac{x}{1 - x ^{2}}

Add a constant to the solution

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

x \to 2 lim (x^{2} - 3 x)

y^{^{''}} + 2 y^{^{'}} = 2 x + 7 - e - 2 x

\int sin (2 x) cos^{4} (2 x) d x

f (x) = (7 x + 6)^{2}

(- 16 x^{2} + 1)^{2}

What is the integral of (sqrt(1-x^2))/((x^2-1)^2) ?
The integral of (sqrt(1-x^2))/((x^2-1)^2) is x/(sqrt(1-x^2))+C