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Integration help

Hi

I have tried so many ways but none actually work. How to solve for this?

∫xx2+2x+7dx\int \frac{x}{x^{2}+2x+7}dx
(edited 10 years ago)
Reply 1
Avatar for BabyMaths
BabyMaths
Original post by Blackberry rocks
Hi

I have tried so many ways but none actually work. How to solve for this?

∫xx2+2x+7dx\int \frac{x}{x^{2}+2x+7}dx


Complete the square in the denominator and use a substitution of u=x+1.

This will allow you to break the integral up into two standard integrals.
Original post by BabyMaths
Complete the square in the denominator and use a substitution of u=x+1.

This will allow you to break the integral up into two standard integrals.



How?

x2+2x+7=(x+1)2+6 x^{2}+2x+7 = (x+1)^2 + 6

let u= x + 1 ; du = dx ; u-1=x

∫u−1u2+6du\int \frac{u-1}{u^{2}+6}du

what then?

I also know that u2+6=(x+1)(x−1)+9∴∫u−1(u+1)(u−1)+7du u^2 + 6 = (x+1)(x-1)+9 \therefore \int \frac{u-1}{(u+1)(u-1)+7}du but it doesn't lead anywhere either
(edited 10 years ago)
Reply 3
Avatar for TenOfThem
TenOfThem
18
Original post by Blackberry rocks
How?

x2+2x+7=(x+1)2+8 x^{2}+2x+7 = (x+1)^2 + 8


You meant +6
Reply 4
Avatar for TenOfThem
TenOfThem
18
Original post by Blackberry rocks
How?

x2+2x+7=(x+1)2+8 x^{2}+2x+7 = (x+1)^2 + 8

let u= x + 1 ; du = dx ; u-1=x

∫u−1u2+8du\int \frac{u-1}{u^{2}+8}du


Can you integrate uu2+6\dfrac{u}{u^2+6}

Can you integrate 1a2+x2\dfrac{1}{a^2+x^2}

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