Re: Stuck on two integrals

Quote:

Originally Posted by

**KevinShaughnessy** Hi,

I've been given two integrals to solve that I can't seem to figure out how to get started on. They are: $\displaystyle\int \frac{x}{\sqrt{x^2 +x + 1}} \, dx$ and $\displaystyle \int \frac{4^x +10^x}{2^x} \, dx$. Could someone help me get started?

for the second one

$\large \dfrac {4^x + 10^x}{2^x}=2^x + 5^x=e^{x\log(2)}+e^{x\log(5)}$

I'll leave you to integrate that, it's easy enough.

Re: Stuck on two integrals

Quote:

Originally Posted by

**KevinShaughnessy** Hi,

I've been given two integrals to solve that I can't seem to figure out how to get started on. They are: $\displaystyle\int \frac{x}{\sqrt{x^2 +x + 1}} \, dx$ and $\displaystyle \int \frac{4^x +10^x}{2^x} \, dx$. Could someone help me get started?

$\displaystyle{\int}\dfrac{x}{\sqrt{x^2+x+1}}~dx =

\displaystyle{\int}\dfrac{(x-\frac 1 2)+\frac 1 2}{\sqrt{(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx=$

$\displaystyle{\int}\dfrac{(x-\frac 1 2)}{\sqrt{(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx +

\displaystyle{\int}\dfrac{\frac 1 2}{\sqrt{(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx$

the first integral make the sub

$u=(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2$

the second one you can match the template in an integral table

$\displaystyle{\int}\dfrac 1 {\sqrt{x^2+a^2}}=\sinh^{-1}\left(\dfrac x a\right)$

Re: Stuck on two integrals

Quote:

Originally Posted by

**romsek** $\displaystyle{\int}\dfrac{x}{\sqrt{x^2+x+1}}~dx =

\displaystyle{\int}\dfrac{(x-\frac 1 2)+\frac 1 2}{\sqrt{(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx=$

$\displaystyle{\int}\dfrac{(x-\frac 1 2)}{\sqrt{(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx +

\displaystyle{\int}\dfrac{\frac 1 2}{\sqrt{(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx$

the first integral make the sub

$u=(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2$

the second one you can match the template in an integral table

$\displaystyle{\int}\dfrac 1 {\sqrt{x^2+a^2}}=\sinh^{-1}\left(\dfrac x a\right)$

That's creative, I like it, thanks. My only problem is that $(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 = x^2 - x + 1$ and not $x^2 +x +1$ no?

For problems like this in the future, is there a method for finding those factors in the denominator or do you just figure it out through intuition or trial and error? I'll probably run into this stuff on a test soon so I would really appreciate knowing that.

Re: Stuck on two integrals

Quote:

Originally Posted by

**KevinShaughnessy** That's creative, I like it, thanks. My only problem is that $(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 = x^2 - x + 1$ and not $x^2 +x +1$ no?

For problems like this in the future, is there a method for finding those factors in the denominator or do you just figure it out through intuition or trial and error? I'll probably run into this stuff on a test soon so I would really appreciate knowing that.

No. You're right. The sub should have been

$u=(x+\frac 1 2)^2 + (\frac {\sqrt{3}}{ 2})^2$

sometimes I whip through these too fast and too late at night

Re: Stuck on two integrals

Quote:

Originally Posted by

**romsek** $\displaystyle{\int}\dfrac{x}{\sqrt{x^2+x+1}}~dx =

\displaystyle{\int}\dfrac{(x-\frac 1 2)+\frac 1 2}{\sqrt{(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx=$

$\displaystyle{\int}\dfrac{(x-\frac 1 2)}{\sqrt{(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx +

\displaystyle{\int}\dfrac{\frac 1 2}{\sqrt{(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx$

the first integral make the sub

$u=(x-\frac 1 2)^2+(\frac {\sqrt 3} 2)^2$

the second one you can match the template in an integral table

$\displaystyle{\int}\dfrac 1 {\sqrt{x^2+a^2}}=\sinh^{-1}\left(\dfrac x a\right)$

Corrected

$\displaystyle{\int}\dfrac{x}{\sqrt{x^2+x+1}}~dx =

\displaystyle{\int}\dfrac{(x+\frac 1 2)-\frac 1 2}{\sqrt{(x+\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx=$

$\displaystyle{\int}\dfrac{(x+\frac 1 2)}{\sqrt{(x+\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx -

\displaystyle{\int}\dfrac{\frac 1 2}{\sqrt{(x+\frac 1 2)^2+(\frac {\sqrt 3} 2)^2}}~dx$

the first integral make the sub

$u=(x+\frac 1 2)^2+(\frac {\sqrt 3} 2)^2$

the second one you can match the template in an integral table

$\displaystyle{\int}\dfrac 1 {\sqrt{x^2+a^2}}=\sinh^{-1}\left(\dfrac x a\right)$