1. ## Intergrate

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.

2. Originally Posted by Bust2000
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 $\int \frac {2x - 3}{x^2 + 16}~dx = \int \frac {2x}{x^2 + 16}~dx - \int \frac 3{x^2 + 16}~dx$

for the first, do a substitution $u = x^2 + 16$

for the second, do not complete the square. instead, factor out the 16 from the denominator and then do a substitution $t = \frac x4$, and go for the arctangent integral

3. ## Question

Can you give me a little deminstration of what you are saying you are saying do a u substition of intergral uv -intergral v du. Sorry I need to work with my equation converter more it doesnt recognize intergrals still. Thanks

4. Originally Posted by Bust2000
Can you give me a little deminstration of what you are saying you are saying do a u substition of intergral uv -intergral v du.
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

Sorry I need to work with my equation converter more it doesnt recognize intergrals still. Thanks
$\int \frac 3{x^2 + 16}~dx = \frac 3{16} \int \frac 1{\frac {x^2}{16} + 1}~dx = \frac 3{16} \int \frac 1{\left( \frac x4 \right)^2 + 1} ~dx$

now, does that integral look familiar? it should. now do a substitution $t = \frac x4$ to see.

5. Hello, Bust2000!

Intergrate: . $
\int\frac{2x-3}{x^2+16}\, dx$
Make two integrals: . $\int\frac{2x\,dx}{x^2+16} \;-\;3\int\frac{dx}{x^2+16}$

The first integral is:. $\int\frac{2x\,dx}{x^2+16}$

. . Let $u \,=\,x^2+16\quad\Rightarrow\quad du\,=\,2x\,dx$

. . Substitute: .[ $\int\frac{du}{u} \;=\;\ln|u| + C\quad\Rightarrow\quad \ln(x^2+16) + C$

And the second integral is of the form: . $\int\frac{du}{u^2+a^2}\;=\;\frac{1}{a}\arctan\left (\frac{u}{a}\right) + C$