I have this (it looks simple) example
$$\int \frac{3}{2-5x}\,dx$$
It looks really simple, no logarithms or trigonometric functions, but I just cannot get the proper result here...
Asked
Active
Viewed 48 times
2
-
1Try: $t=2-5x. $ – borg May 12 '14 at 19:07
-
1A right result is $-\frac{3}{5}\ln(|2-5x|)+C$. But there are equivalent ways to write it. What did you get? – André Nicolas May 12 '14 at 19:08
-
set $u=2-5x$... – Fermat May 12 '14 at 19:08
1 Answers
1
\begin{align} u & = 2 - 5x \\[15pt] du & = -5\,dx \\[15pt] -du/5 & = dx \\[15pt] \int \frac{3}{2-5x}\,dx & = \int \frac{3}{u} \left(\frac{du}{-5}\right) \\[15pt] & =\frac{-3}{5}\int \frac 1 u \, du \\[15pt] & = \frac{-3}{5}\log_e |u| + C \\[15pt] & = \frac{-3}{5}\log_e|2-5x| + C \end{align}