What is the integral of ((2x+1))/(3(1-x-x^2)^2) ?

The integral of ((2x+1))/(3(1-x-x^2)^2) is 1/(3(-x^2-x+1))+C

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integral of ((2x+1))/(3(1-x-x^2)^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= \frac{1}{3} \cdot \int \frac{8 u}{( - u ^{2} + 5 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot 8 \cdot \int \frac{u}{( - u ^{2} + 5 ) ^{2}} d u

Apply u-substitution

= \frac{1}{3} \cdot 8 \cdot \int - \frac{1}{2 v ^{2}} d v

Take the constant out:

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

= \frac{1}{3} \cdot 8 (- \frac{1}{2} \cdot \int \frac{1}{v ^{2}} d v)

Apply the Power Rule

= \frac{1}{3} \cdot 8 (- \frac{1}{2} (- \frac{1}{v}))

Substitute back

= \frac{1}{3} \cdot 8 (- \frac{1}{2} (- \frac{1}{- ( 2 x + 1 ) ^{2} + 5}))

Simplify

\frac{1}{3} \cdot 8 (- \frac{1}{2} (- \frac{1}{- ( 2 x + 1 ) ^{2} + 5})) : \frac{1}{3 ( - x ^{2} - x + 1 )}

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

Add a constant to the solution

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

What is the integral of ((2x+1))/(3(1-x-x^2)^2) ?
The integral of ((2x+1))/(3(1-x-x^2)^2) is 1/(3(-x^2-x+1))+C