Integration by substitution (2x^5+3x^2)/(1+2x^3)^(1/2)

[T_TEngineer_AdamT_T]

i got stuck with this substitution problem:

\(\displaystyle \L \int\, \frac{2x^5\, +\, 3x^2}{\sqrt{1\, +\, 2x^3\, }}\, dx\)

Let u = 1 + 2x^3
du = 6x^2 dx

\(\displaystyle \L \int\, {\frac{x^2\, (2x^3\, +\, 3)}{\sqrt{u\, }}\, \frac{du}{6x^2}\)

\(\displaystyle \L \frac{1}{6}\, \int\, {\frac{2\left(\frac{u}{2}\, -\, \frac{1}{2}\right)\, +\, 3}{\sqrt{u\, }}\, du\)

\(\displaystyle \L \frac{1}{6}\, \int\, \frac{u\, +\, 2}{\sqrt{u\, }}\, du\)

omg and i got stuck with this part which does not match the answer key w/c is:

\(\displaystyle \L \frac{1}{9}\, \sqrt{1\, +\, 2x^3\, }\, \left(2x^3\, +\, 7\right)\, +\, C\)

i think my substitution is wrong... help me

 

[galactus]

Hey Adam:

Rewrite as:

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

Then, let \(\displaystyle \L\\u=2x^{3}+3, \;\ \frac{du}{6}=x^{2}dx, \;\ u-3=2x^{3}\)

This gives:

\(\displaystyle \L\\\frac{1}{6}\int\frac{u}{\sqrt{u-2}}du\)

Now, it'll fall together.

 

[T_TEngineer_AdamT_T]

thanks so much galactus!

 

[soroban]

Hello, T_TEngineer_AdamT_T!

Your substitution would have worked . . .

\(\displaystyle \L \int\, \frac{2x^5\, +\, 3x^2}{\sqrt{1\, +\, 2x^3}}\,dx \;=\;\int\frac{x^2(2x^3\,+\,3)}{\sqrt{1\,+\,2x^3}}\,dx\)

Let \(\displaystyle u \:= \:1\,+\,2x^3\;\;\Rightarrow\;\;du\:=\:6x^2\,dx\;\;\Rightarrow\;\;dx \,=\,\frac{du}{6x^2}\)
. . Also: \(\displaystyle \:2x^3\,+\,3\:=\:u\,+\,2\)

Substitute: \(\displaystyle \L\:\int\, {\frac{x^2\, (u\,+\,2)}{\sqrt{u}}\, \frac{du}{6x^2} \;=\;\frac{1}{6}\int\frac{u\,+\,2}{u^{\frac{1}{2}}} du \;=\;\frac{1}{6}\int\left(u^{\frac{1}{2}}\,+\,2u^{-\frac{1}{2}}\right)\,dx\)

. . \(\displaystyle \L=\;\frac{1}{6}\left(\frac{2}{3}u^{\frac{3}{2}}\,+\,4u^{\frac{1}{2}}\right)\,+\,C \;=\;\frac{1}{9}u^{\frac{1}{2}}(u\,+\,6)\,+\,C\)


Back-substitute:

. . \(\displaystyle \L\frac{1}{9}\left(2x^3\,+\,1\right)^{\frac{1}{2}}\left(2x^3\,+\,1\,+\,6\right)\,+\,C \;=\;\frac{1}{9}\left(2x^3\,+\,1)^{\frac{1}{2}}\left(2x^3\,+\,7\right) \,+\,C\)

 

[T_TEngineer_AdamT_T]

thanks soroban!