Question

how to solve: integation root of (3x+4)^3 dx.pls help...

Answer 1

this is the Q:\[\int\limits \sqrt{(3x + 4)^3 dx}\]

Answer 2

make u substitution , for u=3x+4

Answer 3

so now eq will be \[\frac{ 1 }{ 3 }\ \int\limits \sqrt{u^3du}\]

Answer 4

what is final ans?

Answer 5

yeah that would be the integral now you can rewrite that as
\[\frac{ 1 }{3 }\int\limits_{}^{}u^{\frac{ 3 }{ 2 }}du\]

Answer 6

cool wow u did it.... thannnnnk uuuuuuu. so much :)

Answer 7

no problem ur welcome :)

Answer 8

i got ans as \[1/4*\log u ^{5\div6}+c\]

Answer 9

Q is :\[\int\limits dx \div \sqrt[3]{(4x-3)^\frac{ 5 }{ 2 }}\]

Answer 10

i got above ans for this Q. ans is right?

Answer 11

give me a min to check

Answer 12

you got mistake
\[\int\limits_{}^{}{\frac{ du }{ u^{5/6} }}\]
is not
\[\log{u^{\frac{ 5 }{ 6 }}}\]

Answer 13

when you get to that integral you rewrite it as
\[\int\limits_{}^{}u^{-\frac{ 5 }{ 6 }}du\]
and then you solve it with the formula
\[\int\limits_{}^{}x^{n}dx=\frac{ x^{n+1} }{ n+1 }\]

Answer 14

ok got it.... silly mistake there. thanks once again...

Answer 15

no problem ^^

Answer 16

why is \[\int\limits \frac{ dx }{ 1+x^2 }=\tan^{-1} x=-\cos^{-1}x\]

Answer 17

how tan−1x=−cos−1x in above formula?

Answer 18

hm i am not sure if that is correct , never came across it

Answer 19

Q is:\[\int\limits \frac{ dx }{ \sqrt{1-36x^2} }\]

Answer 20

ok so you can rewrite that as
\[\int\limits\limits_{}^{}\frac{ dx }{ \sqrt{36(\frac{ 1 }{ 36 }-x^{2}}) }=\frac{ 1 }{ 6 } \int\limits\limits_{}^{}\frac{ dx }{ \sqrt{(\frac{ 1 }{ 6 })^{2}-x^{2}} }\]
now using table integral
\[\int\limits_{}^{}\frac{ dx }{ \sqrt{a^{2}-x^{2}} }=\sin^{-1}(\frac{ x }{ a })\]
yu can solve it :)

Answer 21

yes solved it ty :)