HELP with intagration by parts

[T_TEngineer_AdamT_T]

Integrate this:
\(\displaystyle \L\\\int{\frac{x^3}{\sqrt{1-x^2}}dx\)

im getting confused with this problem
after integrating by parts:
\(\displaystyle \L\\\frac{x^4}{4\sqrt{1-x^2}}-\int{\frac{x^4}{4}(\frac{x}{(1-x^2)^{\frac{3}{2}}\)
thats where i got stuck

the answer at the back of my book should be:
\(\displaystyle \L\\-x^2\sqrt(1-x^2)-\frac{2}{3}(1-x^2)^{\frac{3}{2}}+C\)

Wahh w8 just figuring out how to use the latex here

 

[galactus]

Integrate this:
\(\displaystyle \L\\\int{\frac{x^3}{\sqrt{1-x^2}}dx\)

im getting confused with this problem
after integrating by parts:
\(\displaystyle \L\\\frac{x^4}{4\sqrt{1-x^2}}-\int{\frac{x^4}{4}\left(\frac{x}{(1-x^2)^{\frac{3}{2}}\right)\)
thats where i got stuck

the answer at the back of my book should be:
\(\displaystyle \L\\-x^2\sqrt(1-x^2)-\frac{2}{3}(1-x^2)^{\frac{3}{2}}+C\)

Wahh w8 just figuring out how to use the latex here

Hello Adam. Get tired of the LaTex free MathHelpForum?.

On this site you must enclose in \(\displaystyle instead of $$Do you have to use parts or can you use another any other method?.$$\)

 

[pka]

Please edit your post so as to correct the LaTeX
If you type \(\displaystyle \L\:\int{\frac{x^3}{\sqrt{1-x^2}}dx \)

You will get \(\displaystyle \L\:\int{\frac{x^3}{\sqrt{1-x^2}}dx\)

 

[T_TEngineer_AdamT_T]

hey thanks...
yes its what the lesson is all about but not sure if this' right
but you need to use int by parts

 

[pka]

Here are the parts: \(\displaystyle \L u = x^2 \quad \& \quad dv = \frac{{xdx}}{{\sqrt {1 - x^2 } }}.\)

 

[T_TEngineer_AdamT_T]

Here are the parts: \(\displaystyle \L u = x^2 \quad \& \quad dv = \frac{{xdx}}{{\sqrt {1 - x^2 } }}.\)

hey thanks alot :twisted:

 

[soroban]

Hello, T_TEngineer_AdamT_T!

\(\displaystyle \L \int \frac{x^3}{\sqrt{1-x^2}}dx\)

\(\displaystyle \begin{array}{ccccccc}u & \,=\, & x^2 & \;\;\; & dv & \,=\, & x(1\,-\,x^2)^{-\frac{1}{2}}dx \\
du & = & 2x\,dx & \;\;\; & v & = & -(1\,-\,x^2)^{\frac{1}{2}} \end{array}\)


We have: \(\displaystyle \L\:-x^2(1\,-\,x^2)^{\frac{1}{2}} \,-\,\int(1\,-\,x^2)^{\frac{1}{2}}(-2x\,dx)\)

. . \(\displaystyle \L= \;-x^2(1\,-\,x^2)^{\frac{1}{2}} \,- \,\frac{2}{3}(1\,-\,x^2)^{\frac{3}{2}} \,+\,C\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I'm surprised that the answer was left like that.
. . Traditionally, textbooks simplify answers beyond all recognition.

Factor: \(\displaystyle \L\:-\frac{1}{3}(1\,-\,x^2)^{\frac{1}{2}}\,\left[3x^2\,+\,2(1\,-\,x^2)\right] \,+\,C\)

. . . .\(\displaystyle \L=\;-\frac{1}{3}(1\,-\,x^2)^{\frac{1}{2}}(x^2\,+\,2)\,+\,C\)