Question

Evaluate the integral.
(x-3)/x dx from 1 to 4

Answer 1

rewrite the problem by splitting the numerator

\[\int\limits_{1}^{4} \frac{x}{x} - \frac{3}{x} dx = \int\limits_{1}^{4} 1 - \frac{3}{x} dx\]

Answer 2

For the integrand substitute u=x-3 and du=dx
\[\int\limits_{1}^{4}\frac{ u }{ u+3 }du\]
Do long division:
\[\int\limits\limits_{1}^{4}1-\frac{3}{ u+3 }du\]

Integrate term by term:
\[\int\limits_{1}^{4}1du-3\int\limits_{1}^{4}\frac{ 1 }{ u+3 }du\]

Substitute s=u+3 and ds=du
\[\int\limits_{1}^{4}1du-3\int\limits_{1}^{4}1/sds\]

Integral of 1/s is log(s)
so we get
u-3log(s)
sub back in for u
(x-3)log(x)-3
By plugging in the limits of integration we get:
3-3log(4)

Answer 3

its not that difficult a question ... it doesn't need substitution