Question

I really need help on this. Take integral of (×^2)(×+2)^(1/3)

Answer 1

you can try the substitution \[y^3=(x+2)\]

Answer 2

I tried u substitution but du doesn't equal something in the problem so I couldn't easily substitute

Answer 3

\[\Large \int\limits x^2 \sqrt[3]{\color{orangered}{x+2}}\;dx\]

U sub will work :) It's just a little tricky.
We want to make the substitution, \(\large \color{orangered}{u=x+2}\).
Taking the derivative of our substitution gives us, \(\large du=dx\)

Answer 4

\[\Large \int\limits x^2 \sqrt[3]{\color{orangered}{x-2}}\;dx \qquad\to\qquad \int\limits x^2 \sqrt[3]{\color{orangered}{u}}\;du\]So we still need to deal with the x^2.
Where can we get an x^2 from? Hmm let's look back at our substitution and solve it for x^2.\[\large u=x+2\]Subtracting 2 from each side,\[\large x=u-2\]Squaring both sides,\[\large x^2=(u-2)^2\]There it is! Now we can replace all of the x's with u's! :)\[\large \int\limits (u-2)^2\cdot u^{1/3}\;du\]

Answer 5

From here you can expand out the square, multiply and then apply the power rule to each term.

Answer 6

Omg!!!!!!!! Thanks sooooo much!!!

Answer 7

I copied it down aand everything u said makes ssense!!!

Answer 8

oh good \c:/

Answer 9

\[\large \int \color{blue}x^2\color{red}{(x+2)}^{1/3}\color{green}{dx}\]
let \[\large \color{red}{y^3=x+2\\\large\color{blue}{y^3-2=x}\\\large \color{green}{3y^2dy=dx} }\]
\[\large \int \color{blue}x^2\color{red}{(x+2)}^{1/3}\color{green}{dx}\\\large\quad =\int\color{blue}{(y^3-2)}^2\color{red}{(y^3)}^{1/3}\color{green}{(3y^2dy)}\]
this integral is radical-free

Answer 10

How do u get y raised to power 3?

Answer 11

Let I=\[\int\limits x ^{2}\sqrt[3]{x-2} dx\]
\[put \left( x-2 \right)^{\frac{ 1 }{ 3 }}=u,x-2=u ^{3},x=u ^{3}+2,dx=3 u ^{2} du\]
\[I=\int\limits \left( u ^{3}+2 \right)^{2}u*3u ^{2}du=\int\limits 3u ^{3}\left( u ^{6}+4u ^{3}+4 \right)du\]
\[I=3\int\limits \left( u ^{9}+4u ^{6}+4u ^{3} \right) du=3\left( \frac{ u ^{10} }{ 10 }+\frac{ 4u ^{7} }{7 }+\frac{ 4u ^{4} }{4 } \right)+c\]
Replace the value of u and get the answer.

Answer 12

sorry i wrote x-2 in place of x+2

Answer 13

\[after correction I=3\left( \frac{ u ^{10} }{ 10 } -\frac{ 4u ^{7} }{7 }+\frac{ 4u ^{4} }{ 4 }\right)+c\]
Replace u=x+2

Answer 14

\[Replace u=\left( x+2 \right)^{\frac{ 1 }{3 }}\]
\[I=3\left( \frac{ \left( x+2 \right)^{\frac{ 10 }{ 3 }} }{10 } -\frac{ 4\left( x+2 \right)^{\frac{ 7 }{3 }} }{7 }+\left( x+2 \right)^{\frac{ 4 }{3 }}\right)+c\]