Question

integral (1/(pix -4)^2 )

I want to know how to answer this question?

Answer 1

what's pix?

Answer 2

pi times x

Answer 3

\[\int\frac1{(\pi x-4)^2}dx\]
Is this correct?

Answer 4

Yes

Answer 5

sub u=(πx−4) and du= π dx

Answer 6

Let:
u = pi * x-4
du = pi dx
\[\therefore \pi^{-1}\int u^{-2}du=-\pi*u^{-1}+C=-\frac{1}{\pi u}+C\]
plug back in:
\[\frac1{4\pi-\pi^2x}+C\]

Answer 7

i would use substitution
u=πx−4
du=π dx

so this would be
\[ 1/\pi \int\limits 1/ u ^{2} du \]
the integral of 1/u^2 is -1/u

so you get
\[1/4\pi- \pi^2x + C\]

Substitute back for u = pi x-4:
= 1/(4 pi-pi^2 x)+constant

Answer 8

I don't understand how do you get that 1/pi before the integral?

Answer 9

it comes from the substitution