Question

Evaluate the integral 23/x^3-125 dx
I've worked this about half way and am stuck.
I've found A/x-5 + Bx+C/x^2+5x+25=
A(x^2+5x+25)+(Bx+C)(x-5)
I also found A=1/75, B= -1/75, C= -4/25
Then:
Integral 23/x^3-125 dx = 23Integral (1/75)/x-5 dx +23Integral (-1/75)x-(4/25)/x^2+5x+25 dx
= 23/75 ln |x-5|-23/75Integral x+12/x^2+5x+25
And this is where I get lost. :(

Answer 1

Hold on, I think you may have made a slight error...\[\int\frac{23}{x^3-125}~dx=\int\frac{23}{(x-5)(x^2+5x+25)}~dx\]
\[\begin{align*}\frac{23}{(x-5)(x^2+5x+25)}&=\frac{A}{x-5}+\frac{Bx+C}{x^2+5x+25}\\
23&=Ax^2+5Ax+25A+Bx^2+Cx-5Bx-5C\\
23&=(A+B)x^2+(5A-5B+C)x+25A-5C
\end{align*}\]
\[\begin{cases}A+B=0\\5A-5B+C=0\\25A-5C=23\end{cases}~~\Rightarrow~~A=\frac{23}{75},~B=-\frac{23}{75},~C=-\frac{46}{15}\]

Answer 2

Thank you, yes, I see that. I'm still stuck at the same place, however.
I have \[\frac{ 23 }{ 75 }\ln \left| x-5 \right|-\frac{ 23 }{ 75 }\int\limits_{}^{}\frac{ x+10 }{x^2+5x+25 }\]
and I'm not sure what to do with that integral now.
Thanks in advance for anyone's help! :)

Answer 3

Oops I forgot the dx after that integral...
\[\frac{ 23 }{ 75 }\ln \left| x-5 \right|-\frac{ 23 }{ 75 }\int\limits\limits_{}^{}\frac{ x+10 }{x^2+5x+25 } dx\]

Answer 4

Notice that letting \(u=x^2+5x+25\), you get \(du=(2x+5)~dx\). What I would do first is to split up the fraction by rewriting the numerator so that you can make use of this substitution:
\[\frac{x+10}{x^2+5x+5}=\frac{1}{2}\cdot\frac{2x+20}{x^2+5x+25}=\frac{1}{2}\left(\frac{2x+5}{x^2+5x+5}+\frac{15}{x^2+5x+5}\right)\]
So now,
\[\frac{23}{75}\ln|x-5|-\frac{23}{75}\int\frac{x+10}{x^2+5x+5}~dx\]
is equal to
\[\frac{23}{75}\ln|x-5|-\frac{23}{150}\int\frac{2x+5}{x^2+5x+5}~dx-\frac{23}{10}\int\frac{dx}{x^2+5x+5}\]
Substituting, you get
\[\frac{23}{75}\ln|x-5|-\frac{23}{150}\int\frac{du}{u}~dx-\frac{23}{10}\int\frac{dx}{x^2+5x+5}\]
For the remaining integral, complete the square:
\[\frac{1}{x^2+5x+5}=\frac{1}{x^2+5x+\dfrac{25}{4}-\dfrac{5}{4}}=\frac{1}{\left(x+\dfrac{5}{2}\right)^2-\dfrac{5}{4}}\]
Then you would substitute \(x+\dfrac{5}{2}=\dfrac{\sqrt5}{2}\sec t\), so you get
\[\frac{23}{75}\ln|x-5|-\frac{23}{150}\int\frac{du}{u}~dx-\frac{23}{10}\int\frac{dx}{\left(x+\dfrac{5}{2}\right)^2-\dfrac{5}{4}}\]
\[\frac{23}{75}\ln|x-5|-\frac{23}{150}\int\frac{du}{u}~dx-\frac{23}{10}\int\frac{\dfrac{\sqrt5}{2}\sec t\tan t}{\left(\dfrac{\sqrt5}{2}\sec t\right)^2-\dfrac{5}{4}}~dt\]
\[\frac{23}{75}\ln|x-5|-\frac{23}{150}\int\frac{du}{u}~dx-\frac{23\sqrt5}{20}\int\frac{\sec t\tan t}{\dfrac{5}{4}\sec^2 t-\dfrac{5}{4}}~dt\]
\[\frac{23}{75}\ln|x-5|-\frac{23}{150}\int\frac{du}{u}~dx-\frac{23\sqrt5}{25}\int\frac{\sec t\tan t}{\sec^2 t-1}~dt\]
\[\frac{23}{75}\ln|x-5|-\frac{23}{150}\int\frac{du}{u}~dx-\frac{23\sqrt5}{25}\int\frac{\sec t\tan t}{\tan^2 t}~dt\]
\[\frac{23}{75}\ln|x-5|-\frac{23}{150}\int\frac{du}{u}~dx-\frac{23\sqrt5}{25}\int\frac{\sec t}{\tan t}~dt\]
\[\frac{23}{75}\ln|x-5|-\frac{23}{150}\int\frac{du}{u}~dx-\frac{23\sqrt5}{25}\int\csc t~dt\]

Answer 5

Thank you so much!!