Question

use substitution to find the indefinite integral..

Answer 1

\[\int\limits_{}^{} \left( \frac{ 6 }{\left( 6x+1 \right)^{2} } \right) dx\]

Answer 2

u = (6x+1)
du = 6dx

Answer 3

\(u=6x+1\), so \(\dfrac{1}{6}du=dx\):
\[\frac{1}{6}\int\frac{6}{u^2}~du\]

Answer 4

The derivative of 6x +1 is 6.

Answer 5

ok so sorry I'm slow .. i have to times \[\frac{ 1 }{6}\] by \[\frac{ 6 }{ u ^{2} }\] rights?

Answer 6

Actually in this question they cancel out.. so you're left with just

\[\int\limits_{}^{}du/u^2 \]

Answer 7

so the answer would be \[\frac{ 1 }{ (6x+1)^2} + c\] ?

Answer 8

u^(-2) is a power rule of integration ...

Answer 9

k u^n derives to nk u^(n-1)

n-1 = -2, so n=-1

nk = 1; -k = 1, k = -1

if we need to unfold it that much ....

Answer 10

\[u^{-2}\to-u^{-1}+c\]
or with the actual parts ....
\[ -(6x+1)^{-1}+ c\]

Answer 11

ok i kinda follow but where did you get the -2?

Answer 12

property of exponents:
\[\frac{1}{a^b}=a^{-b}\]

Answer 13

since we have a power rule for integration, its just simpler to have it in that form

Answer 14

\[\int\limits_{}^{} \frac{ e^x-e ^{-x} }{ e^x+e^{-x} } dx\]

Answer 15

Easy substitution: \(u=e^x+e^{-x}\), then \(du=\left(e^x-e^{-x}\right)~dx\), so
\[\int\frac{e^x-e^{-x}}{e^x+e^{-x}}~dx=\int\frac{du}{u}\]

Answer 16

1st. thanks for replying didn't think anyone would lol

2nd. since du/u that means its ln (e^x + x^-x) + C ?

Answer 17

@sithsAndGiggles

Answer 18

yes