Question

I am trying to take this integral (x^2)sqrt(10+25x^2)dx, a form (u^2)sqrt(u^2+-a^2) = (1/8)(u(2u^2+-a^2)sqrt(u^2+-a^2)-a^4lnabs(u+sqrt(u^2+-a^2)), and the fact that I can reduce my "a" after I divide the denominator of 25 out from under the U or "x^2" is confusing me on whether I will make all of my a's (2/5) or (10/25). Basically I think I am supposed to make them 2/5 and pull a 5 out front, but I my professor doesn't give us enough tries online to figure out an algebra snafu. My answer is (5/8)(x(2x^2+(2/5))sqrt(x^2+(2/5))-(4/25)lnabs(x+sqrt(x^2+2/5))....

Answer 1

you are trying to integrate: \[\int\limits_{}^{}x^2 \sqrt{10+25x^2} dx\]?

Answer 2

yes

Answer 3

\[I=\int\limits x^2\sqrt{10+25x^2}dx\]
put \[x=\frac{ 1 }{ t },dx=\frac{ -1 }{ t^2 }dt\]
\[=\int\limits \frac{ 1 }{ t^2 }\sqrt{\frac{ 10 t^2+25 }{ t^2 }}*\frac{ -1 }{ t^2 }dt\]
\[=-\int\limits \frac{ \sqrt{10 t^2+25} ~t }{ t^6 } dt\]
put \[\sqrt{10 t^2+25}=u,10 t^2+25=u^2,20 t~dt=2 ~u~du\]
complete it.

Answer 4

\[\int\limits_{}^{}x^2 \sqrt{10+25x^2} dx=\int\limits_{}^{}x^2\sqrt{10(1+\frac{25 x^2}{10}) } dx\\ =\sqrt{10} \int\limits_{}^{}x^2 \sqrt{1+(\frac{5x}{\sqrt{10}})^2} dx\]
I would choose the sub
\[\tan(\theta)=\frac{5x}{\sqrt{10}}\]
\[\sec^2(\theta) d \theta =\frac{5}{\sqrt{10}} dx \]