Question

Evaluate the integral:
int (x^4 + 6x^3 + 10x^2 + x) dx / (x^2 + 6x +10)

After dividing, I get
int x^2 dx + int (x) dx / (x^2 + 6x +10) and this is where I am stuck...

Answer 1

i don't get your answer.. \[\huge \int x^2 dx + \frac{\int xdx}{x^2 + 6x + 10}\]
????

Answer 2

@xlegendx,
\[\int\frac{x^4+6x^3+10x^2+x}{x^2+6x+10}~dx\]
Finding the quotient via long division yields
\[\int x^2~dx+\int\frac{x}{x^2+6x+10}~dx\]

Answer 3

Ahh, LD..

Answer 4

yes sith... I don't understand how to take care of the second integral now...

Answer 5

lol that makes more sense

Answer 6

lol sorry

Answer 7

if it were me, i would use algebraic substitution since the denominator seems to be prime

Answer 8

I think we'll have to use logarithms for the second part....just a hunch.

Answer 9

substitution:

so u = x^2 + 6x + 10
du = 2x + 6 dx
but I have x not 2x + 6 right?

Answer 10

then just subtract 6 from the numerator so nothing changes

Answer 11

@gadav478, have you tried rewriting the integrand?
\[\int\frac{x}{x^2+6x+10}~dx\]
Note that letting \(u=x^2+6x+10\) gives you \(du=(2x+6)~dx\).
So let's rewrite the integral:
\[\frac{1}{2}\int\frac{2x+6-6}{x^2+6x+10}~dx\\
\frac{1}{2}\left[ \int\frac{2x+6}{x^2+6x+10}~dx - 6\int\frac{dx}{x^2+6x+10}~dx \right]\]
Now applying that substitution:
\[\frac{1}{2}\int\frac{du}{u} - 3\int\frac{dx}{x^2+6x+10}~dx\]
For the second integral, complete the square in the denominator, then use a trig sub.

Answer 12

see?

Answer 13

ohhh

Answer 14

Where did this 1/2 outside the integrand come from....

Answer 15

to neutralize 2x

Answer 16

@Jhannybean, I multiplied the integral by 2/2, then distributed the numerator's 2.

Answer 17

thanks so much everyone

Answer 18

Hmm... i'll go over this.

Answer 19

You're welcome

Answer 20

for anyone still here I have: x^3/3 + ln(x^2 +6x+10) - 6 arctan (x+3) + c

Answer 21

forgot to distribute the 1/2: x^3/3 + ln(x^2 +6x+10) - 3 arctan (x+3) + c