intergrate (2x-3)/(x^2+16) dx
Question is intergrate by partial fractions if so what am I not seeing complete the square in the denominator? Thanks for Checking out question people.
ah, i don't think partial fractions is the way to go here.
note that $\displaystyle \int \frac {2x - 3}{x^2 + 16}~dx = \int \frac {2x}{x^2 + 16}~dx - \int \frac 3{x^2 + 16}~dx$
for the first, do a substitution $\displaystyle u = x^2 + 16$
for the second, do not complete the square. instead, factor out the 16 from the denominator and then do a substitution $\displaystyle t = \frac x4$, and go for the arctangent integral
you are mixing this up with integration by parts, that is not what i am talking about. i am saying you must do integration by substitution. please look up this method, and try not to confuse the two
$\displaystyle \int \frac 3{x^2 + 16}~dx = \frac 3{16} \int \frac 1{\frac {x^2}{16} + 1}~dx = \frac 3{16} \int \frac 1{\left( \frac x4 \right)^2 + 1} ~dx$Sorry I need to work with my equation converter more it doesnt recognize intergrals still. Thanks
now, does that integral look familiar? it should. now do a substitution $\displaystyle t = \frac x4$ to see.
Hello, Bust2000!
Make two integrals: .$\displaystyle \int\frac{2x\,dx}{x^2+16} \;-\;3\int\frac{dx}{x^2+16} $Intergrate: .$\displaystyle
\int\frac{2x-3}{x^2+16}\, dx $
The first integral is:.$\displaystyle \int\frac{2x\,dx}{x^2+16}$
. . Let $\displaystyle u \,=\,x^2+16\quad\Rightarrow\quad du\,=\,2x\,dx$
. . Substitute: .[$\displaystyle \int\frac{du}{u} \;=\;\ln|u| + C\quad\Rightarrow\quad \ln(x^2+16) + C$
And the second integral is of the form: .$\displaystyle \int\frac{du}{u^2+a^2}\;=\;\frac{1}{a}\arctan\left (\frac{u}{a}\right) + C$