Question

hi, have couple integral problems need to be answered using U-substitution, can someone help me?

Answer 1

1) \[\int\limits_{}^{}\frac{ 2x-1 }{ \sqrt{3x ^{2}-3x} } dx\]

Answer 2

i know the answer of this one but i dont know the steps to answer it correctly

Answer 3

Notice that derivative of \(3x^2-3x=6x-3=3(2x-1)\)
so you substitution could be \(u=3x^2-3x\) and \(du=3(2x-1)dx\)
So the integral will become:
\(1/3\int\huge \frac{du}{\sqrt{u}}\)

Answer 4

so is my answer correct? this is what i came up with after getting the antiderivative and substitution :\[\frac{ 2 }{ 3 } \sqrt{3x ^{2} - 3x} + c\]

Answer 5

Yes you are right

Answer 6

Thanks @myko , how about this one: \[\int\limits_{}^{} x^{3} \sqrt{x ^{2}+1}\]
\[u= x ^{2} +1 ?\]

Answer 7

you could go like this: \(u= x^2\), and \(du=2xdx\)
so integral becomes:
\(\int u\sqrt{u+1}du\)
now make one more substitution: \(v=u+1\) and \(dv=du\)
integral becomes:
\(\int (v-1)\sqrt{v}dv=\int (v^{3/2}-v^{1/2})dv\) which is straight forward

Answer 8

sry forgot 1/2 infront of integral after 1º subsitution. Because du=2xdx

Answer 9

i wrote a long solution then the browser crashed :( anyway, did the problem and came out with this answer \[\frac{ 1 }{ 5 } ( x ^{2}+1)^{\frac{ 5 }{ 2 }} - \frac{ 2 }{ 3 }(x ^{2}+1)^{\frac{ 3 }{ 2 }} + c\]

Answer 10

i guess , to lasy to check :). Looks similar, just that 2/3 i tink should be 1/3

Answer 11

how about this one:
\[\int\limits_{1}^{9} (\sqrt{x}-2)dx\]

Answer 12

what's the problema with this.
\(\int_1^9x^{1/2}dx-\int_1^92dx\)

Answer 13

what "u" will be?

Answer 14

you dont need it

Answer 15

4/3?

Answer 16

i guess

Answer 17

thank you