What is the integral of (x^3) / [Sqrt(1+4x^2)]?
- Thread starter chadsgirl
- Start date
Hello, chadsgirl!
This can also be done by parts . . .
\(\displaystyle \begin{array}{ccccccc}u & \,=\, & x^2 & \;\;\; &dv &\,=\, & x(1\,+\,4x^2)^{-\frac{1}{2}}\,dx \\
du & = & 2x\,dx & \;\;\; & v & = & \frac{1}{4}(1\,+\,4x^2)^{\frac{1}{2}}
\end{array}\)
And we have: \(\displaystyle \L\:\frac{1}{4}x^2(1\,+\,4x^2)^{\frac{1}{2}}\,-\,\frac{1}{2}\int x(1\,+\,4x^2)^{\frac{1}{2}}dx\)
. . . \(\displaystyle \L=\;\frac{1}{4}x^2(1\,+\,4x^2)^{\frac{1}{2}}\,-\,\frac{1}{24}(1\,+\,4x^2)^{\frac{3}{2}}\,+\,C\)
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This can be simplified . . .
Factor: \(\displaystyle \L\:\frac{1}{24}(1\,+\,4x^2)^{\frac{1}{2}}\,\cdot\,\left[6x^2\,-\,(1\,+\,4x^2)\right]\,+\,C\)
. . . \(\displaystyle \L=\;\frac{1}{24}(1\,+\,4x^2)^{\frac{1}{2}}(2x^2\,-\,1)\,+\,C\)
This can also be done by parts . . .
\(\displaystyle \begin{array}{ccccccc}u & \,=\, & x^2 & \;\;\; &dv &\,=\, & x(1\,+\,4x^2)^{-\frac{1}{2}}\,dx \\
du & = & 2x\,dx & \;\;\; & v & = & \frac{1}{4}(1\,+\,4x^2)^{\frac{1}{2}}
\end{array}\)
And we have: \(\displaystyle \L\:\frac{1}{4}x^2(1\,+\,4x^2)^{\frac{1}{2}}\,-\,\frac{1}{2}\int x(1\,+\,4x^2)^{\frac{1}{2}}dx\)
. . . \(\displaystyle \L=\;\frac{1}{4}x^2(1\,+\,4x^2)^{\frac{1}{2}}\,-\,\frac{1}{24}(1\,+\,4x^2)^{\frac{3}{2}}\,+\,C\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
This can be simplified . . .
Factor: \(\displaystyle \L\:\frac{1}{24}(1\,+\,4x^2)^{\frac{1}{2}}\,\cdot\,\left[6x^2\,-\,(1\,+\,4x^2)\right]\,+\,C\)
. . . \(\displaystyle \L=\;\frac{1}{24}(1\,+\,4x^2)^{\frac{1}{2}}(2x^2\,-\,1)\,+\,C\)
skeeter
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\(\displaystyle \L u = 1+4x^2\)
\(\displaystyle \L x^2 = \frac{u-1}{4}\)
\(\displaystyle \L du = 8x dx\)
\(\displaystyle \L \int \frac{x^3}{\sqrt{1+4x^2}}dx =\)
\(\displaystyle \L \frac{1}{8} \int x^2 \cdot \frac{8x}{\sqrt{1+4x^2}} dx\)
substitute ...
\(\displaystyle \L \frac{1}{8} \int \frac{u-1}{4} \cdot \frac{1}{\sqrt{u}} du =\)
\(\displaystyle \L \frac{1}{32} \int \sqrt{u} - \frac{1}{\sqrt{u}} du\)
\(\displaystyle \L \frac{1}{32} \left(\frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}}\right) + C\)
\(\displaystyle \L \frac{u^{\frac{1}{2}}}{16} \left(\frac{u}{3} - 1\right) + C\)
\(\displaystyle \L \frac{\sqrt{1+4x^2}}{16} \left(\frac{4x^2-2}{3}\right) + C\)
\(\displaystyle \L \frac{\sqrt{1+4x^2}}{8} \left(\frac{2x^2-1}{3}\right) + C\)
\(\displaystyle \L \frac{\sqrt{1+4x^2}(2x^2-1)}{24} + C\)
\(\displaystyle \L x^2 = \frac{u-1}{4}\)
\(\displaystyle \L du = 8x dx\)
\(\displaystyle \L \int \frac{x^3}{\sqrt{1+4x^2}}dx =\)
\(\displaystyle \L \frac{1}{8} \int x^2 \cdot \frac{8x}{\sqrt{1+4x^2}} dx\)
substitute ...
\(\displaystyle \L \frac{1}{8} \int \frac{u-1}{4} \cdot \frac{1}{\sqrt{u}} du =\)
\(\displaystyle \L \frac{1}{32} \int \sqrt{u} - \frac{1}{\sqrt{u}} du\)
\(\displaystyle \L \frac{1}{32} \left(\frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}}\right) + C\)
\(\displaystyle \L \frac{u^{\frac{1}{2}}}{16} \left(\frac{u}{3} - 1\right) + C\)
\(\displaystyle \L \frac{\sqrt{1+4x^2}}{16} \left(\frac{4x^2-2}{3}\right) + C\)
\(\displaystyle \L \frac{\sqrt{1+4x^2}}{8} \left(\frac{2x^2-1}{3}\right) + C\)
\(\displaystyle \L \frac{\sqrt{1+4x^2}(2x^2-1)}{24} + C\)