evaluate integral of sqrt[1 + sqrt[1 + sqrt[x]]] dx
- Thread starter Smily
- Start date
Hello, Smily!
I think I did it . . . but someone check my work, please!
u\,-\,1)^2\:=\:u^2\,-\,2u\,+\,1\)
\(\displaystyle \;\;\;\sqrt{x}\:=\:u^2\,-\,2u\;\;\Rightarrow\;\;x \:=\u^2\,-\,2u)^2\:=\:u^4\,-\,4u^3\,+\,4u^2\)
Then: \(\displaystyle \,dx\:=\4u^3\,-\,12u^2\,+\,8u)du\:=\:4(u^3\,-\,3u^2 \,+\,2u)\,du\)
Substitute: \(\displaystyle \L\,\int u^{\frac{1}{2}}\cdot4(u^3\,-\,3u^2\,+\,2u)\,du \;= \;4\L\int\left(u^{\frac{7}{2}}\,-\,3u^{\frac{5}{2}}\,+\,2u^{\frac{3}{2}}\right)\,du\)
And we get: \(\displaystyle \L\,4\left(\frac{2}{9}u^{\frac{9}{2}}\,-\,\frac{6}{7}u^{\frac{7}{2}}\,+\.\frac{4}{5}u^{\frac{5}{2}}\right)\,+\,C \;= \;\frac{8}{315}u^{\frac{5}{2}}\left(35u^2\,-\,135u\,+\,126\right)\,+\,C\)
And I'll you back-substitute . . .
I think I did it . . . but someone check my work, please!
Let: \(\displaystyle 1\,+\,\sqrt{1\,+\,\sqrt{x}}\:=\:u\;\;\Rightarrow\;\;\sqrt{1\,+\,\sqrt{x}}\:=\:u\,-\,1\;\;\Rightarrow\;\;1\,+\, \sqrt{x}\:=\
u\,-\,1)^2\:=\:u^2\,-\,2u\,+\,1\)\(\displaystyle \;\;\;\sqrt{x}\:=\:u^2\,-\,2u\;\;\Rightarrow\;\;x \:=\
u^2\,-\,2u)^2\:=\:u^4\,-\,4u^3\,+\,4u^2\)Then: \(\displaystyle \,dx\:=\
4u^3\,-\,12u^2\,+\,8u)du\:=\:4(u^3\,-\,3u^2 \,+\,2u)\,du\)Substitute: \(\displaystyle \L\,\int u^{\frac{1}{2}}\cdot4(u^3\,-\,3u^2\,+\,2u)\,du \;= \;4\L\int\left(u^{\frac{7}{2}}\,-\,3u^{\frac{5}{2}}\,+\,2u^{\frac{3}{2}}\right)\,du\)
And we get: \(\displaystyle \L\,4\left(\frac{2}{9}u^{\frac{9}{2}}\,-\,\frac{6}{7}u^{\frac{7}{2}}\,+\.\frac{4}{5}u^{\frac{5}{2}}\right)\,+\,C \;= \;\frac{8}{315}u^{\frac{5}{2}}\left(35u^2\,-\,135u\,+\,126\right)\,+\,C\)
And I'll you back-substitute . . .
Here's my attempt. Let's see if it agrees with Soroban's.
\(\displaystyle \L\\\int{\sqrt{1+\sqrt{1+\sqrt{x}}}}dx\)
Let \(\displaystyle x=u^{2}\) and \(\displaystyle 2udu=dx\)
\(\displaystyle \L\\2\int{\sqrt{1+\sqrt{1+u}}}udu\)
Let \(\displaystyle 1+u=w^{2}\) and \(\displaystyle du=2wdw\)
\(\displaystyle \L\\4\int\sqrt{1+w}(w^{2}-1)wdw\)
Let \(\displaystyle 1+w=s^{2}\) and \(\displaystyle dw=2sds\)
I'll let you do the subbing to get:
\(\displaystyle \L\\\int(16s^{4}-24s^{6}+8s^{8})ds\)
Now, integrating and reverting back to terms of x, we end up with(I'll leave the details to you):
\(\displaystyle \L\\\frac{16}{5}(1+\sqrt{1+\sqrt{x}})^{\frac{5}{2}}+\frac{8}{9}(1+\sqrt{1+\sqrt{x}})^{\frac{9}{2}}-\frac{24}{7}(1+\sqrt{1+\sqrt{x}})^{\frac{7}{2}}\)
Check it all out. I do believe it jives with Soroban's solution, just a different format.
\(\displaystyle \L\\\int{\sqrt{1+\sqrt{1+\sqrt{x}}}}dx\)
Let \(\displaystyle x=u^{2}\) and \(\displaystyle 2udu=dx\)
\(\displaystyle \L\\2\int{\sqrt{1+\sqrt{1+u}}}udu\)
Let \(\displaystyle 1+u=w^{2}\) and \(\displaystyle du=2wdw\)
\(\displaystyle \L\\4\int\sqrt{1+w}(w^{2}-1)wdw\)
Let \(\displaystyle 1+w=s^{2}\) and \(\displaystyle dw=2sds\)
I'll let you do the subbing to get:
\(\displaystyle \L\\\int(16s^{4}-24s^{6}+8s^{8})ds\)
Now, integrating and reverting back to terms of x, we end up with(I'll leave the details to you):
\(\displaystyle \L\\\frac{16}{5}(1+\sqrt{1+\sqrt{x}})^{\frac{5}{2}}+\frac{8}{9}(1+\sqrt{1+\sqrt{x}})^{\frac{9}{2}}-\frac{24}{7}(1+\sqrt{1+\sqrt{x}})^{\frac{7}{2}}\)
Check it all out. I do believe it jives with Soroban's solution, just a different format.