Can someone help me ?
topsquark Senior Member Joined Aug 27, 2012 Messages 2,370 Jun 17, 2021 #2 This is a moderately bad one. I'll give you the steps to get you started and you fill in the blanks. ∫80(x2+2x+10)2 dx\int \dfrac{80}{(x^2 + 2x + 10)^2} ~ dx∫(x2+2x+10)280 dx Step 1: Complete the square: x2+2x+10=(x+1)2+9x^2 + 2x + 10 = (x + 1)^2 + 9x2+2x+10=(x+1)2+9. Step 2: Let u=x+1u = x + 1u=x+1 Step 3: Let u=3 tan(v)u = 3 ~ tan(v)u=3 tan(v). This gives ∫1(x2+2x+10)2 dx=127∫cos2(v) dv\int \dfrac{1}{(x^2 + 2x + 10)^2} ~ dx = \dfrac{1}{27} \int cos^2(v) ~ dv∫(x2+2x+10)21 dx=271∫cos2(v) dv Fill in the steps and you can take the integration from there. Let us know if you need more help and show what you've done. -Dan Addendum: You could probably do this with partial fractions. I haven't looked at it that way yet.
This is a moderately bad one. I'll give you the steps to get you started and you fill in the blanks. ∫80(x2+2x+10)2 dx\int \dfrac{80}{(x^2 + 2x + 10)^2} ~ dx∫(x2+2x+10)280 dx Step 1: Complete the square: x2+2x+10=(x+1)2+9x^2 + 2x + 10 = (x + 1)^2 + 9x2+2x+10=(x+1)2+9. Step 2: Let u=x+1u = x + 1u=x+1 Step 3: Let u=3 tan(v)u = 3 ~ tan(v)u=3 tan(v). This gives ∫1(x2+2x+10)2 dx=127∫cos2(v) dv\int \dfrac{1}{(x^2 + 2x + 10)^2} ~ dx = \dfrac{1}{27} \int cos^2(v) ~ dv∫(x2+2x+10)21 dx=271∫cos2(v) dv Fill in the steps and you can take the integration from there. Let us know if you need more help and show what you've done. -Dan Addendum: You could probably do this with partial fractions. I haven't looked at it that way yet.