Question

integral help

Answer 1

Partial fractions, know what that is ?

Answer 2

for the first question? I can use partial fractions straight away?

Answer 3

infact partial fractions may not be even required.
just a u substitution can solve it

Answer 4

the denominator is a perfect square, isn't it ?
of what ?

Answer 5

(x+1)^2

Answer 6

yes, so you could just do
u = x+1

x= ..
du = ...?

Answer 7

du = dx?

Answer 8

correct
and x = u-1

now plug these into your original integral to make a new integral

Answer 9

wait sorry, why does x = u-1, where does the u-1 come from?

Answer 10

i would help but im clueless

Answer 11

u = x+1

x= .... ?

Answer 12

because we will need 'x' in terms of 'u' for the numerator

Answer 13

ok

Answer 14

so whats your new integral ?

Answer 15

integral of 7(u-1)+13 / u^2 ?

Answer 16

correct
is that easy for you to solve ?

Answer 17

sorry, I'm not good at any of differentiation or integral problems. Would you try split it in two terms of something?

Answer 18

thats correct

(7u+6)/u^2 = 7/u + 6/u^2

then use x^n formula for 2nd term
and 1/x formula for the first term, you have the formula list ?

Answer 19

can you just find the integral of 7u^-1 and 6u^-2?

Answer 20

i can yes
i'll give you the formula so that you can too
\(\int (1/x)dx = \ln |x|+c \\ \int x^n dx = \dfrac{x^{n+1}}{n+1}+c\)

Answer 21

for 2nd term, since we have u^-2
plug in n=-2

Answer 22

ok, so I'm not sure if i'm doing it correctly at all but would you get log|u| - 1/u + 2C?

Answer 23

what about constants 7 and 6?

Answer 24

and then you substitute your u=x+1 if that's correct I believe

Answer 25

the 7 will remain in front of log|u|? and 6 will remain in front of -1/u ?