Integration by substitution: int x^3 (x^2 + 1)^3/2 dx
- Thread starter cmnalo
- Start date
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,204
\(\displaystyle \L \int x^3 (x^2+1)^{\frac{3}{2}} dx\)
\(\displaystyle \L u = x^2 + 1\)
\(\displaystyle \L x^2 = u - 1\)
\(\displaystyle \L du = 2x dx\)
\(\displaystyle \L \frac{1}{2} \int 2x \cdot x^2 (x^2+1)^{\frac{3}{2}} dx\)
\(\displaystyle \L \frac{1}{2} \int (u-1)u^{\frac{3}{2}} du\)
\(\displaystyle \L \frac{1}{2} \int u^{\frac{5}{2}} - u^{\frac{3}{2}} du\)
take it from here?
\(\displaystyle \L u = x^2 + 1\)
\(\displaystyle \L x^2 = u - 1\)
\(\displaystyle \L du = 2x dx\)
\(\displaystyle \L \frac{1}{2} \int 2x \cdot x^2 (x^2+1)^{\frac{3}{2}} dx\)
\(\displaystyle \L \frac{1}{2} \int (u-1)u^{\frac{3}{2}} du\)
\(\displaystyle \L \frac{1}{2} \int u^{\frac{5}{2}} - u^{\frac{3}{2}} du\)
take it from here?
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,204
multiply it out for yourself ...
(1/2)*2x*x<sup>2</sup> = x<sup>3</sup>
had to break it up ... you only need one x for du (2x dx = du)
the leftover x<sup>2</sup> can be defined in terms of u ...
since u = x<sup>2</sup> + 1, x<sup>2</sup> = u - 1
then do the substitution.
(1/2)*2x*x<sup>2</sup> = x<sup>3</sup>
had to break it up ... you only need one x for du (2x dx = du)
the leftover x<sup>2</sup> can be defined in terms of u ...
since u = x<sup>2</sup> + 1, x<sup>2</sup> = u - 1
then do the substitution.
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,204
\(\displaystyle \L (u - 1)u^{\frac{3}{2}}\)
distribute the \(\displaystyle \L u^{\frac{3}{2}}\) ...
\(\displaystyle \L u \cdot u^{\frac{3}{2}} - u^{\frac{3}{2}} = u^1 \cdot u^{\frac{3}{2}} - u^{\frac{3}{2}}\)
so ... what do you get when you simplify the first term?
distribute the \(\displaystyle \L u^{\frac{3}{2}}\) ...
\(\displaystyle \L u \cdot u^{\frac{3}{2}} - u^{\frac{3}{2}} = u^1 \cdot u^{\frac{3}{2}} - u^{\frac{3}{2}}\)
so ... what do you get when you simplify the first term?
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,204
you're having a time with the algebra, aren't you?
\(\displaystyle \L \frac{1}{2} \left(\frac{2}{7}u^{\frac{7}{2}} - \frac{2}{5}u^{\frac{5}{2}} \right) + C\)
now ... factor out \(\displaystyle \L 2u^{\frac{5}{2}}\) from the terms in ( ) ...
\(\displaystyle \L \frac{1}{2} \cdot 2u^{\frac{5}{2}} \left(\frac{u}{7} - \frac{1}{5} \right) + C\)
\(\displaystyle \L u^{\frac{5}{2}} \left(\frac{5u-7}{35} \right) + C\)
back-substitute \(\displaystyle \L x^2 + 1\) for \(\displaystyle \L u\) ...
\(\displaystyle \L (x^2 + 1)^{\frac{5}{2}} \left[\frac{5(x^2+1)-7}{35}\right] + C\)
\(\displaystyle \L (x^2 + 1)^{\frac{5}{2}} \left(\frac{5x^2-2}{35}\right) + C\)
\(\displaystyle \L \frac{1}{35} (x^2 + 1)^{\frac{5}{2}} (5x^2-2) + C\)
\(\displaystyle \L \frac{1}{2} \left(\frac{2}{7}u^{\frac{7}{2}} - \frac{2}{5}u^{\frac{5}{2}} \right) + C\)
now ... factor out \(\displaystyle \L 2u^{\frac{5}{2}}\) from the terms in ( ) ...
\(\displaystyle \L \frac{1}{2} \cdot 2u^{\frac{5}{2}} \left(\frac{u}{7} - \frac{1}{5} \right) + C\)
\(\displaystyle \L u^{\frac{5}{2}} \left(\frac{5u-7}{35} \right) + C\)
back-substitute \(\displaystyle \L x^2 + 1\) for \(\displaystyle \L u\) ...
\(\displaystyle \L (x^2 + 1)^{\frac{5}{2}} \left[\frac{5(x^2+1)-7}{35}\right] + C\)
\(\displaystyle \L (x^2 + 1)^{\frac{5}{2}} \left(\frac{5x^2-2}{35}\right) + C\)
\(\displaystyle \L \frac{1}{35} (x^2 + 1)^{\frac{5}{2}} (5x^2-2) + C\)