# Thread: Integral and series help.

1. ## Integral and series help.

1. Integral of (x(ln(x))/(square root of (x^2)-1) dx

2. Integral of cos^2(1+ln(x))/x dx

3. Evaluate the following sum: (SERIES QUESTION)

n=1 to infinity (3^(n+2))/(2^(2n-1))

2. Originally Posted by devil10078
1. Integral of (x(ln(x))/(square root of (x^2)-1) dx

2. Integral of cos^2(1+ln(x))/x dx

3. Evaluate the following sum: (SERIES QUESTION)

n=1 to infinity (3^(n+2))/(2^(2n-1))
#3.

$\displaystyle \displaystyle 2^{(2n-1)}={{2^{2n}}\over{2}}={{4^{n}}\over{2}}$.

$\displaystyle \displaystyle 3^{(n+2)}=9\left(3^n\right)$.

So $\displaystyle \displaystyle {{3^{(n+2)}}\over{2^{(2n-1)}}}= {{9\left(3^n\right)}\over{{{4^{n}}\over{2}}}}= {{18\left(3^n\right)}\over{4^{n}}}$.

Therefore, $\displaystyle \displaystyle \sum_{n=1}^\infty}{{3^{(n+2)}}\over{2^{(2n-1)}}}= \sum_{n=1}^\infty}18{{3^n}\over{4^{n}}}$

Can you take it from there?

3. Originally Posted by devil10078
]
2. Integral of cos^2(1+ln(x))/x dx
Let $\displaystyle u = 1+\ln x$. What does the integral become?

4. $\displaystyle \displaystyle \int \frac{\cos^2(\ln{x}+1)}{x} \, dx$

$\displaystyle u = \ln{x} + 1$

$\displaystyle \displaystyle du = \frac{1}{x} \, dx$

substitute ...

$\displaystyle \displaystyle \int \cos^2{u} \, du = \frac{1}{2} \int 1 + \cos(2u) \, du$

integrate then back sub ...

5. Hello, devil10078!

I pray that these are not on your exam . . .

$\displaystyle \displaystyle (1)\;\;I \;=\;\int\frac{x \ln x }{\sqrt{x^2-1}}\,dx$

Integrate by parts:

. . $\displaystyle \begin{array}{ccccccccc} u &=& \ln x && dv &=& x(x^2-1)^{\frac{1}{2}}\,dx \\ du &=& \dfrac{dx}{x} && v &=& \frac{1}{3}(x^2-1)^{\frac{3}{2}} \end{array}$

$\displaystyle \displaystyle I \;=\;\tfrac{1}{3}(x^2-1)^{\frac{3}{2}}\ln x - \tfrac{1}{3}\int \frac{(x^2-1)^{\frac{3}{2}}}{x}\,dx$

The new integral reguires Trig Substitution:

. . $\displaystyle x \,=\,\sec\theta \quad\Rightarrow\quad dx \,=\,\sec\theta\tan\theta\,d\theta$

Good Luck!