Question

How do I find the indefinite integral (calculus) for:

Answer 1

(attached).

Answer 2

sample problem:

Answer 3

don't be hoodwinked by the \(\sqrt{3x}\) this is the same as \(\sqrt{3}\sqrt{x}\)so start with
\[\frac{2}{\sqrt{3}}\int\frac{dx}{\sqrt{x}}\]and use the power rule backwards

Answer 4

ok i did not read carefully enough but it is the same idea. start with
\[\frac{2}{\sqrt[3]{3}}\int\frac{dx}{\sqrt[3]{x}}\]

Answer 5

or if you prefer
\[\frac{2}{\sqrt[3]{3}}\int x^{-\frac{1}{3}}dx\]

Answer 6

is the answer (3x)^2/3

Answer 7

How?

Answer 8

is it correct or not ??

Answer 9

I don't know. I haven't been given an answer to check myself.

Answer 10

u can do this question by substitution :
let 3x = y^3
differentiating both side wrt x
we have dx=y^2dy
got it!

Answer 11

then substituting the value of dx in original integral we have