## patbatE21 Group Title Evaluate the integral: Indefinite Integral ((x^3 + x^2 + 2x + 1) / (x^2 + 1)(x^2 + 2)) dx 2 years ago 2 years ago

1. patbatE21 Group Title

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2. patbatE21 Group Title

so far I have x^3(A+C) + x^2 (B+D) + x(2A +C) +(2B + D)

3. patbatE21 Group Title

But I believe I'm going in the wrong direction

4. slaaibak Group Title

${{x^3 + x^2 + 2x + 1}\over{(x^2 + 1)(x^2 + 2)} } = {{Ax + B}\over{x^2 + 1}} + {{Cx + D}\over{x^2 + 2}}$ By partial franctions. Now:

5. patbatE21 Group Title

ok

6. patbatE21 Group Title

I'm good so far

7. slaaibak Group Title

$x^3 + x^2 + 2x + 1 = (Ax + B)(x^2 + 2) + (Cx + D)(x^2 + 1)$ $x^3 + x^2 + 2x + 1 = Ax^3 + 2Ax + Bx^2 + 2B + Cx^3 + Cx + Dx^2 + D$ $x^3 + x^2 + 2x + 1 = (A + C)x^3 + (B + D)x^2 + (2A + C)x + (2B + D)$ A + C = 1 B + D = 1 2A + C = 2 2B + D = 1

8. patbatE21 Group Title

Alright...but the answer I saw was (1/2)ln (x^2 + 1) + (1 / square(2)) arctan(x/square(2)) + C

9. patbatE21 Group Title

It's a frustrating problem

10. slaaibak Group Title

Lets take it further then: By solving A, B, C, D: A = 1 C=0 B=0 D=1

11. patbatE21 Group Title

ok

12. slaaibak Group Title

${{x^3 + x^2 + 2x + 1}\over{(x^2 + 1)(x^2 + 2)}} = {x \over {x^2 + 1} } + {{1}\over{x^2 + 2}}$

13. slaaibak Group Title

Now the fun starts! Now we integrate each one of those

14. patbatE21 Group Title

ohhhhh ok

15. slaaibak Group Title

Now to integrate the first one: $\int\limits {x \over {x^2 + 1} } dx$ $x^2 = u$ $2x dx = du$ $xdx ={1\over2}du$ ${1\over2}\int\limits{1\over{u + 1}} du$ =${1\over2}\ln(u + 1) + C$ resubstituting u with x^2: ${1\over2}\ln(x^2 + 1) + C$

16. slaaibak Group Title

Do you need help with the second part of the integral?

17. patbatE21 Group Title

$\int\limits{1\over{x^2 + 2}} dx$ now, by factoring out the 2: $\int\limits {1\over{2({1\over2}x^2 + 1)}} dx$ ${1\over2}\int\limits{1\over{({1\over2}x^2 + 1)}}dx$ Putting the 1/2 inside the bracket: ${1\over2} \int\limits { {1}\over{ ({{x}\over{\sqrt2}})^2 + 1 } }dx$ ${x \over {\sqrt 2} } = u$ $dx = \sqrt2 du$ ${1\over2}\int\limits {{\sqrt2}\over{u^2+1}} du$ ${{\sqrt2}\over2}\int\limits {{du}\over{u^2+1}}$ = ${\sqrt2 \over 2} \arctan u + c$ substituting back: ${\sqrt2 \over 2} \arctan({x \over \sqrt2}) + C$ since ${\sqrt2 \over 2 } = {1\over \sqrt2}$ ${1\over \sqrt2} \arctan({x \over \sqrt2}) + C$