Question

Integrate 2x-1/36+(x-3)^2

Answer 1

2x-1/36(1+(x-3)^2/36)
u=x-3/6 u^2=(x-3)^2/36
12u+5=(2x-6)+5
12u+5/36(1+u^2)

Answer 2

break in to two parts

Answer 3

one will give an arctangent, the other a log

Answer 4

assuming it is \[\int \frac{2x-1}{36+(x-3)^2}dx\]

Answer 5

\[\int\limits_{}^{}12u/36(1+u^2)+\int\limits_{}^{}5/36(1+u^2)\]

Answer 6

and yes its in this form @satellite73

Answer 7

Which integral will result in an inverse tangent function? Can you perform this integration? Which method would you use to integrate the other part of this integral problem?

Answer 8

=1/6ln(1+u^2)+5/36tan^-1(u)

Answer 9

I've checked and found that your first integral is fine.

Answer 10

Next time you want verification of correctness, please show all of your work as part of the deal.

Answer 11

what was your substitution for \(dx\)

Answer 12

Okay I am sorry, but my answer is incorrect according to the answer sheet.

Answer 13

So here is what i did i factored out the 36 from the denominator and used the susbtitution method where u=x-3/6 therefore u^2=(x-3)^2/36. I then rewrote the expression u=x-3/6 to be 12u+5 to cope with the numerator.
12u+5/36(1+u^2) did a split to the fraction and then integrated both
12u/36(1+u^2)+5/36(1+u^2)

Answer 14

if \[u=\frac{x-3}{6}\]

then \[6u=x-3\]

and \[6du=dx\]

Answer 15

Oh thanks I just realised my mistake here i forgot the du part

Answer 16

you also need to replace u in the end. your final answer should be a function of x

Answer 17

Here is my final answer after i have altered some stuff
ln(1+(x-3/6)^2)+5/6tan^-1(x-3/6)

Answer 18

+c