Question

integral sqrt (x^2+64)/ 5x^2
please help

Answer 1

\(\Large\bf \color{#008353}{\text{Welcome to OpenStudy! :)}}\)

Answer 2

\[\Large\bf\sf \int\limits \frac{\sqrt{x^2+64}}{5x^2}dx\]

Answer 3

Ummm so you have a few different options.
A `Trig Sub` is one of them.

Are you familiar with that method?

Answer 4

yes

Answer 5

\[\Large\bf\sf x=8\tan \theta\]Do you understand why I chose that for the substitution?

Answer 6

yes becuase the triangle in my book has us do that i also know that \[dx =8\sec ^2\]

Answer 7

\[\Large\bf\sf dx=8\sec^2\theta \;d \theta\]Ok good.

Answer 8

Plug all the stuff in! :)

Answer 9

thats where I'm getting stuck

Answer 10

\[\Large\bf\sf \int\limits\limits \frac{\sqrt{x^2+64}}{5x^2}dx\quad=\quad \int\limits\limits \frac{\sqrt{64\tan^2\theta+64}}{5\cdot64\tan^2\theta}(8\sec^2\theta\;d \theta)\]

Answer 11

That look ok so far?
Simplifying will be a little tricky :)

Answer 12

\[I=\int\limits \frac{ \sqrt{x^2+64} }{ 5x^2 }dx\]
\[=\int\limits \frac{ \left( x^2+64 \right) }{5x^2\sqrt{x^2+64} }dx\]
\[Put~ x=\frac{ 1 }{ t },dx=-\frac{ 1 }{ t^2 }dt\]
\[I=\int\limits \frac{ \frac{ 1 }{ t^2 }+64 }{ \frac{ 5 }{ t^2 }\sqrt{\frac{ 1 }{ t^2 }+64} }\left( \frac{ -1 }{ t^2 }\right) dt\]
\[=\frac{ 1 }{5 }\int\limits \frac{ -1\left( 1+64t^2 \right) }{t \sqrt{1+64t^2} }dt\]
\[=-\frac{ 1 }{5 }\int\limits \frac{ \left( 1+64t^2 \right)t~dt }{ t^2\sqrt{1+64t^2} }\]
put \[\sqrt{1+64t^2}=y,~64 ~t^2=y^2-1,t^2=\frac{ y^2-1 }{64 },2t~dt=\frac{ 2ydy }{64 }\]
\[tdt=\frac{ ydy }{64 }\]
\[I=-\frac{ 1 }{5}\int\limits \frac{ y^2*\frac{ y~dy }{ 64 } }{\frac{ y^2-1 }{64 }y }=-\frac{ 1 }{5 }\int\limits \frac{ y^2dy }{y^2-1 }\]
\[=-\frac{ 1 }{5 }\int\limits \frac{ y^2-1+1 }{ y^2-1 }dy=-\frac{ 1 }{5 }\left[ \int\limits dy+\int\limits \frac{ 1 }{ (y+1)(y-1) }dy \right]\]
complete it.

Answer 13

make partial fractions and complete.

Answer 14

Also consider doing integration by parts
\[
\text{dv}=\frac{1}{5 x^2}\\

u=\sqrt{x^2+64}\\
v=-\frac 1 {5 x}\\
du=\frac{x}{\sqrt{x^2+64}}\\
Integral=
-\frac{\sqrt{x^2+64}}{5 x}+
\int \frac 1 {5 x} \frac{x}{\sqrt{x^2+64}}dx=\\
-\frac{\sqrt{x^2+64}}{5 x}+
\int \frac 1 {5 } \frac{ dx}{\sqrt{x^2+64}} =\\
\frac{1}{5} \left(-\frac{\sqrt{x^2+64}
}{x}+\sinh
^{-1}\left(\frac{x}{8}\right)\right) +C
\]

Answer 15

@zepdrix

Answer 16

\[\frac{ 1 }{ \left( y+1 \right)\left( y-1 \right) }=\frac{ A }{y+1}+\frac{ B }{y-1 }\]
1=A(y-1)+B(y+1)
put y=1
1=A(1-1)+B(1+1)
B=1/2
put x=-1
1=A(-1-1)+B(-1+1)
A=-1/2
\[\int\limits \frac{ 1 }{\left( y+1 \right)\left( y-1 \right) }dy=\frac{ 1 }{2 } \int\limits \left( \frac{ -1 }{ y+1 }+\frac{ 1 }{y-1 } \right)dy\]
\[=\frac{ 1 }{ 2 }\left( -\ln \left( y+1 \right)+\ln \left( y-1 \right) \right)=-\frac{ 1 }{2 }\ln \frac{ y+1 }{ y-1 }\]
\[I=-\frac{ 1 }{ 5 }\left[ y-\frac{ 1 }{ 2 }\ln \frac{ y+1 }{ y-1 } \right]+c\]
\[I=-\frac{ 1 }{5 }\left[ \sqrt{1+64t^2}-\frac{ 1 }{2 }\ln \frac{ \sqrt{1+64t^2}+1 }{\sqrt{1+64t^2}-1 } \right]+c\]
\[I=\frac{ -1 }{5 }\left[ \sqrt{1+\frac{ 64 }{ x^2 }}-\frac{ 1 }{ 2 }\ln \frac{ \sqrt{1+\frac{
64 }{ x^2 }}+1 }{\sqrt{1+\frac{ 64 }{ x^2 }}-1 } \right]+c\]
\[I=-\frac{ 1 }{5 }\left[ \frac{ \sqrt{x^2+64} }{ x }-\frac{ 1 }{ 2 } \ln\frac{\sqrt{x^2+64} +x }{ \sqrt{x^2+64}-x }\right]+c\]

Answer 17

as suggested by Mr. eliassaab
\[\int\limits \frac{ 1 }{ \sqrt{x^2+64} }dx,\
put~x=8\tan t,dx=8\sec ^2t~dt\]
\[\int\limits \frac{ 8~\sec ^2 ~t~dt }{ 8\sqrt{\tan ^2t+1} }=\int\limits \frac{ \sec ^2t~dt }{ \sec t }=\int\limits \sec t~dt=\ln \left| \sec t+\tan t \right|\]
\[=\ln \left| \tan t+\sqrt{\tan ^2 t+1} \right|=\ln \left| \frac{ x }{ 8 }+\frac{ \sqrt{x^2+64} }{ 8 } \right|\]

Answer 18

\[\frac{ 1 }{ 2 }\ln \frac{ \sqrt{x^2+64}+x }{ \sqrt{x^2+64}-x }\times \frac{ \sqrt{x^2+64}+x }{ \sqrt{x^2+64}+x }\]
\[=\frac{ 1 }{ 2 } \ln \frac{ \left( \sqrt{x^2+64}+x \right)^2 }{ x^2+64-x^2 }\]
\[=\frac{ 1 }{ 2 }\ln \left( \frac{ \sqrt{x^2+64}+x }{ 8 } \right)^2\]
\[=2\times \frac{ 1 }{ 2 }\ln \left( \frac{ \sqrt{x^2+64}+x }{ 8 } \right)=\ln \left( \frac{ \sqrt{x^2+64}+x }{ 8 } \right)\]