I'm stuck with an integration $$\int\left [{2\over x(1+xy)}+{3y\over 1+xy}\right ]\text{d}x$$ Here y is constant. Please help me out. Thanks in advance.
Integrate $\int \left [{2\over x(1+xy)}+{3y\over 1+xy}\right ]dx$
2
$\begingroup$
integration
-
0$y$ is some constant ? – 2017-01-15
-
0yeah, y is constant – 2017-01-15
-
0what is the upper limit of integration? – 2017-01-15
-
0it's an indefinite integral – 2017-01-15
-
0partial fractions for the first, the second is logarithm – 2017-01-15
4 Answers
2
hint: $\dfrac{2}{x(1+xy)}= \dfrac{2}{x} - \dfrac{2y}{1+xy}, \displaystyle \int\dfrac{kydx}{1+xy}= k\ln(1+xy)+C, k = 2,3$ .
1
$\displaystyle \int \frac{2}{x(1+xy)}dx = 2\int \frac{1}{x}+2\int \frac{-y}{y(\frac{1}{y}+x)}$
$\displaystyle \int\frac{3y}{y(\frac{1}{y}+x)}dx$
Now evaluate both of them.
1
Hint： $$\begin{align*} \int \left [ {2\over x(1+xy)}+{3y\over 1+xy} \right ]\mathrm{d}x&=2\int \left ( \frac{1}{x}-\frac{y}{1+xy} \right )\mathrm{d}x+3y\int \frac{1}{1+xy}\, \mathrm{d}x\\ &=2\int \frac{1}{x}\, \mathrm{d}x+y\int \frac{1}{1+xy}\, \mathrm{d}x \end{align*}$$
0
$\displaystyle \int \frac{2 + 3xy}{x(1 + xy)}dx$
$\displaystyle \int \frac{2 + 2xy + xy}{x(1 + xy)}dx$
= $\displaystyle \int \frac{2 + 2xy}{x(1 + xy)} dx + \int \frac{xy}{x(1 + xy)} dx$
= $\displaystyle \int \left[\frac{2}{x} + \frac{y}{(1 + xy)}\right] dx$
= $\displaystyle \int \left[\frac{2}{x} + \int \frac{1}{\frac{1}{y} + x} \right]dx$
= $\displaystyle 2 \log|x| + \log|\frac1y + x| + c$