## anonymous one year ago Help me to demonstrate this integral, please :)

1. anonymous

2. anonymous

it's integration by parts. $\int\limits u~dv=uv-\int\limits v~du$ $u = x^n$ $dv=e^x dx$

3. anonymous

$\int\limits f(x).g(x)dx=f(x).\int\limits g(x) dx-\int\limits \big [f'(x) \int\limits g(x)dx \big ]dx$ This is formula for finding integral of a product of two functions, the process is known as integration as parts you can also exchange f and g $\int\limits f(x).g(x)dx=g(x).\int\limits f(x) dx-\int\limits \big [g'(x) \int\limits f(x)dx \big ]dx$ Out of these 2 formulae, we select the one which is easier and less complicated of course, sometimes it may even be unsolvable one way around This method is also useful for finding integrals of a single function, in these cases we can take second function as a constant function 1 example $\int\limits \ln(x)dx=\int\limits \ln(x).1dx=\ln(x).\int\limits1.dx-\int\limits \big [\frac{1}{x}.\int\limits 1.dx\big]dx$ $\implies \int\limits \ln(x)dx=\ln(x).x-\int\limits[\frac{1}{x}.x]dx=x\ln(x)-\int\limits dx=x\ln(x)-x+C$ You may factor out the x $\int\limits \ln(x)dx=x(\ln|x|-1)+C$ If we take ln x as second function we can't solve it because we don't know the integral of ln x, that's the whole point of that integral We are taking mod inside ln because ln only takes positive values