Question

Question #2ec64

Answer 1

`x^2/2+x-2/sqrt5ln|(2x+1-sqrt5)/(2x+1+sqrt5)|+C.`

Explanation:

Let, `I=int(x^3+2x^2-3)/(x^2+x-1)dx`

By Long Division, we get,

`(x^3+2x^2-3)/(x^2+x-1)=x+1-2/(x^2+x-1),` so,

`I=int(x+1)dx-2int1/(x^2+x-1)dx`

`=x^2/2+x-2int1/{x^2+x+1/4-5/4)dx`

`=x^2/2+x-2int1/{(x+1/2)^2-(sqrt5/2)^2} dx`

But, we know that, `int1/(x^2-a^2)=1/(2a)ln|(x-a)/(x+a)|+c.,`

`:., I=x^2/2+x-2/(2*sqrt5/2)ln|(x+1/2-sqrt5/2)/(x+1/2+sqrt5/2)|,i.e.,`

`I=x^2/2+x-2/sqrt5ln|(2x+1-sqrt5)/(2x+1+sqrt5)|+C.`

Enjoy Maths.!

Answer 2

You can factor using the quadratic formula.

Explanation:

You have done this part:

`int(x^3+2x^2-3)/(x^2+x-1)dx= intxdx+intdx-2int1/(x^2+x-1)dx`

Quadratic formula:

`x = (-b+-sqrt(b^2-4(a)(c)))/(2a)`

`x = (-1+-sqrt(1^2-4(1)(-1)))/(2(1))`

`x = (-1+-sqrt(5))/2`

This makes the factors:

`(x+(1+sqrt5)/2)(x+(1-sqrt5)/2)`

Partial Fraction decomposition:

`1/((x+(1+sqrt5)/2)(x+(1-sqrt5)/2)) = A/(x+(1+sqrt5)/2)+ B/(x+(1-sqrt5)/2)`

Multiply by the denominator:

`1 = A(x+(1-sqrt5)/2)+ B(x+(1+sqrt5)/2)`

Eliminate B by letting `x = -(1+sqrt5)/2`

`1 = A(-(1+sqrt5)/2+(1-sqrt5)/2)`

`1 = A(-sqrt5)`

`A = -sqrt5/5`

Eliminate A by letting `x = -(1-sqrt5)/2`

`1 = B(-(1-sqrt5)/2+(1+sqrt5)/2)`

`1 = B(sqrt5)`

`B = sqrt5/5`

The integral becomes:

`int(x^3+2x^2-3)/(x^2+x-1)dx= intxdx+intdx-2sqrt5/5int1/(x+(1-sqrt5)/2)dx + 2sqrt5/5int1/(x+(1+sqrt5)/2)`

The integrals are trivial:

`int(x^3+2x^2-3)/(x^2+x-1)dx= x^2/2+x-2sqrt5/5ln|x+(1-sqrt5)/2| + 2sqrt5/5ln|x+(1+sqrt5)/2| + C`