# evaluate the following integrals: a)      ∫x5/x4-x3+8x-8 b)      ∫x2/√x2+4x+3

embizze | Certified Educator

(a) Evaluate `int x^5/(x^4-x^3+8x-8)dx` :

Use long division on the integrand to get:

`=int (1+x+(x^3-8x^2+8)/(x^4-x^3+8x-8))dx`

For the last term of the integrand factor the denominator and expand by partial fractions:

`=int (1+x+1/(9(x-1))+8/(9(x+2))-16/(3(x^2-2x+4)))dx`

Integrate term by term:

`=x+x^2/2+1/9ln|x-1|+8/9ln|x+2|-16/3 int (dx)/(x^2-2x+4)`

** `-16/3 int (dx)/(x^2-2x+4)` Complete the square in the denominator:

`=-16/3 int (dx)/((x-1)^2+3)`   This is of the form `int (du)/(u^2+a^2)`

`=-16/3 [1/sqrt(3)"arctan"(x-1)/sqrt(3)]`

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`int x^5/(x^4-x^3+8x-8)dx`

`=x+x^2/2+1/9ln|x-1|+8/9ln|x+2|-16/3[1/sqrt(3)"arctan"(x-1)/sqrt(3)]+C` ------------------------------------------------------------------

(b) `intx^2/(sqrt(x^2+4x+3))dx` Complete the square in the radicand:

`=int x^2/sqrt((x+2)^2-1)dx`  Let u=x+2; then du=dx and x=u-2:

`=int (u-2)^2/sqrt(u^2-1)du` Let u=sec(s); then du=sec(s)tan(s)ds and `(du)/tan(s)=secsds` : ** Note that `sec^2(s)-1=tan^2(s)` **

`=int(sec(s)-2)^2sec(s)ds` Expand:

`=int (4-4sec(s)+sec^2(s))sec(s)ds`

`=4intsec(s)ds-4int sec^2(s)ds+int sec^3(s)ds`

`=4ln|sec(s)+tan(s)|-4tan(s)+int sec^3(s)ds`

** `intsec^3(s)ds` Use the reduction formula

`int sec^m(s)ds=(sec^(-1+m)(s)sin(s))/(-1+m)+(-2+m)/(-1+m)int sec^(-2+m)(s)ds` So `int sec^3(ds)`

`=(sec^2(s)sin(s))/2+1/2int sec(s)ds`

`=1/2 sec(s)tan(s)+1/2ln|sec(s)+tan(s)|`

So we have:

`4ln|sec(s)+tan(s)-4tan(s)|+1/2sec(s)tan(s)+1/2ln|sec(s)+tan(s)|` Substitute sec(s)=u and `tan(s)=sqrt(u^2-1)` to get:

`=4ln|u+sqrt(u^2-1)|-4sqrt(u^2-1)+1/2 usqrt(u^2-1)+1/2ln|u+sqrt(u^2-1)|` Substitute u=x+2:

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

`+1/2(x+2)sqrt((x+2)^2-1)+1/2ln|x+2+sqrt((x+2)^2-1)|`

Combining like terms we get:

`-4sqrt(x^2+4x+3)+1/2(x+2)sqrt(x^2+4x+3)`

`+9/2ln|x+2+sqrt(x^2+4x+3)|`