Question

find the integral of [(x^2)divided by squareroot of (x+2)] dx

Answer 1

\[\int \frac{x^2}{\sqrt{x+2}}dx\]

Answer 2

is that Right ?

Answer 3

yes

Answer 4

hmm you can add 4 and negative 4 in numerator hmm let me do this

Answer 5

\[ \frac{x^2 +4 -4 }{\sqrt{x+2}} \implies \frac{(x+2)(x-2)}{\sqrt{x+2}} + \frac{4}{\sqrt{x+2}}\]

Answer 6

sorry no medal.. I need the way on how to answer this

Answer 7

oh don't worry about that Lol

Answer 8

hmm I have a better way now put \( x+2 = t^2\)

Answer 9

\[x = t^2 - 2 \]
\[dx = 2x* dt \]
\[\frac{x^2}{\sqrt{t^2}}\implies \frac{(t^2-2)^2}{t}\]

Answer 10

Now you can do it

Answer 11

Correction :\[dx = 2t* dt\]

Answer 12

\[2\int \frac{\cancel{t}(t^4 + 4 - 4t) }{\cancel t}dt\]

Answer 13

\[2 \int( t^4 +4 -4t^2)dt\]