Need help with 2 Integrals

**purpleclover** · April 13th 2012, 09:40 AM

Could u guys give me some tips about how to aproach these integrals please? http://i284.photobucket.com/albums/l...hreat/Mate.png (Img in attachment too.) Thanks Alot!

**Prove It** · April 13th 2012, 09:53 AM

Originally Posted by purpleclover

Could u guys give me some tips about how to aproach these integrals please? http://i284.photobucket.com/albums/l...hreat/Mate.png (Img in attachment too.) Thanks Alot!

$\displaystyle \begin{align*} \frac{(x + 1)^3}{(x + 2)(x + 3)(x + 4)^2} = \frac{x^3 + 3x^2 + 3x + 1}{(x + 2)(x + 3)(x + 4)^2} \end{align*}$

Now apply partial fractions with $\displaystyle \begin{align*} \frac{A}{x + 2} + \frac{B}{x + 3} + \frac{C}{(x + 4)} + \frac{D}{(x + 4)^2} = \frac{x^3 + 3x^2 + 3x + 1}{(x + 2)(x +3)(x + 4)^2} \end{align*}$

**purpleclover** · April 13th 2012, 09:56 AM

Thanks alot!... How about the second one ?

**Prove It** · April 13th 2012, 10:03 AM

Originally Posted by purpleclover

Could u guys give me some tips about how to aproach these integrals please? http://i284.photobucket.com/albums/l...hreat/Mate.png (Img in attachment too.) Thanks Alot!

$\displaystyle \begin{align*} \int{\frac{x+1}{\sqrt{x^2 + x + 1}}\,dx} &= \frac{1}{2}\int{\frac{2x + 2}{\sqrt{x^2 + x + 1}}\,dx} \\ &= \frac{1}{2}\int{\frac{2x + 1 + 1}{\sqrt{x^2 + x + 1}}\,dx} \\ &= \frac{1}{2}\int{\frac{2x + 1}{\sqrt{x^2 + x + 1}} + \frac{1}{\sqrt{x^2 + x + 1}}\,dx} \\ &= \frac{1}{2}\int{ \frac{2x+1}{\sqrt{x^2 + x + 1}} + \frac{1}{ \sqrt{ x^2 + x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 + 1} }\,dx} \\ &= \frac{1}{2}\int{\frac{2x+1}{\sqrt{x^2+x+1}} + \frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^2 + \frac{3}{4}}}\,dx} \end{align*}$

The first term can be integrated by making the substitution $\displaystyle \begin{align*} u = x^2 + x + 1 \end{align*}$ and the second can be integrated by making the substitution $\displaystyle \begin{align*} x + \frac{1}{2} = \frac{\sqrt{3}}{2}\sinh{t} \end{align*}$ .

**purpleclover** · April 13th 2012, 10:06 AM

holy C***....

THANKS ALOT!

**purpleclover** · April 13th 2012, 10:30 AM

i was woundering in the second integral instead of making the substituion x+1/2=sinht... cant i just make the substituion x+1=t...and then use the ln|..| formula?

**Prove It** · April 13th 2012, 10:38 AM

Originally Posted by purpleclover

i was woundering in the second integral instead of making the substituion x+1/2=sinht... cant i just make the substituion x+1=t...and then use the ln|..| formula?

No you can't. The ln|...| formula only works when the top is the derivative of the bottom, which it isn't.

**purpleclover** · April 13th 2012, 10:39 AM

k thanks.

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