# I need some helps in integrations !!! Really URGENT

• October 30th 2008, 05:43 PM
nns91
I need some helps in integrations !!! Really URGENT
integrate:

1. 1/ x^2* sqr(x^2+4) dx sqr is square root

2. e^(2x)*sin(e^x)dx ( I know I should do by parts but I am stucked)

3. 1/ sqr(x)*(x^(1/3)+1)dx

4. x^(2)+1 / (x^(3)-4x^2)dx

5. 1/ (x* sqr(9+4*x^(2)))dx

6.x^(2)/ sqr(2x-x^(2)) dx

7. cost / (2 -(sint)^(2)) dt ( I guess it's an exact derivative here but I can't move on)

8. x^(3)/ (a^(2)-x^2) dx I can't move on with trig substitution

9. lnx / (x+1)^2 dx

10. e^(ax) * cos(bx) dx By Parts ? how can I do it ?

11. x* e^(sqr(x))dx

12. Arctan (sqr(x)) dx

13. (tan(x))^6 dx

14. (sec(x))^6dx

15. x^(1/3) / (1+x) dx

Some more to come !!

Please give me some help. I really need it.
• October 30th 2008, 05:58 PM
Mathstud28
Quote:

Originally Posted by nns91
integrate:

1. 1/ x^2* sqr(x^2+4) dx sqr is square root

2. e^(2x)*sin(e^x)dx ( I know I should do by parts but I am stucked)

3. 1/ sqr(x)*(x^(1/3)+1)dx

4. x^(2)+1 / (x^(3)-4x^2)dx

Some more to come !!

Please give me some help. I really need it.

1. Let $x=\frac{1}{u}$
2. Let $e^x=u$ and realize that $e^{2x}=e^x\cdot{e^x}$

3. Let $x=u^6$

4.Partial fraction decomp.
• October 30th 2008, 06:06 PM
nns91
Can you explain more detail please ?
• October 30th 2008, 06:08 PM
Mathstud28
Quote:

Originally Posted by nns91
Can you explain more detail please ?

Which ones in particular are you having difficulties with?
• October 30th 2008, 06:12 PM
nns91
All actually, but can we start with the second one first ?
• October 30th 2008, 06:32 PM
Mathstud28
Quote:

Originally Posted by nns91
All actually, but can we start with the second one first ?

$\int{e^{2x}\sin(e^x)}dx=\int{e^x\cdot{e^x}\sin\lef t(e^x\right)dx}$

If we let $u=e^x\Rightarrow{du=e^xdx}$

So out integral becomes

$\int{e^u\sin(u)}du$

Which can be easily computed with two iterations of integration by parts
• October 30th 2008, 06:34 PM
nns91
I thought e^2x = u^2 ?

how about the others ?
• October 30th 2008, 06:35 PM
Mathstud28
Quote:

Originally Posted by nns91
integrate:

5. 1/ (x* sqr(9+4*x^(2)))dx

6.x^(2)/ sqr(2x-x^(2)) dx

7. cost / (2 -(sint)^(2)) dt ( I guess it's an exact derivative here but I can't move on)

8. x^(3)/ (a^(2)-x^2) dx I can't move on with trig substitution

9. lnx / (x+1)^2 dx

10. e^(ax) * cos(bx) dx By Parts ? how can I do it ?

Some more to come !!

Please give me some help. I really need it.

5. Let $2x=3\tan(\theta)$

6. Complete the square on the bottom and let $x+1=u$

7. Let $u=\sin(x)\Rightarrow{du=\cos(x)dx}$

8. Use polynomial divison first

9. Integrate by parts...you could $\ln(x)=u$ first if you were so inclined

10. yes parts...just remember that the a and b are constants
• October 30th 2008, 06:42 PM
nns91
7. That's how I started but the n I got in the problem of the integrating 1/2-u^(2)
• October 30th 2008, 06:43 PM
Mathstud28
Quote:

Originally Posted by nns91
7. That's how I started but the n I got in the problem of the integrating 1/2-u^(2)

$2-u^2=\left(\sqrt{2}-u\right)\left(\sqrt{2}+u\right)$

Use P-fracs
• October 30th 2008, 06:51 PM
nns91
Do you mind explain a little bit more detail about each one ?

I am kinda weak in P fraction so How can I decompose number 4 ?
• October 30th 2008, 06:57 PM
Mathstud28
Quote:

Originally Posted by nns91
Do you mind explain a little bit more detail about each one ?

I am kinda weak in P fraction so How can I decompose number 4 ?

Sorry but it is way past my bedtime...If no one has finished these I will do so tomorrow.
• October 30th 2008, 07:49 PM
Soroban
Hello, nns91!

Quote:

$1)\;\;\int \frac{dx}{x^2\sqrt{x^2+4}}$
Let $x = 2\tan\theta\quad\Rightarrow\quad dx = 2\sec^1\theta\,d\theta$

Substitute: . $\int\frac{2\sec^2\!\theta\,d\theta}{4\tan^2\!\thet a\cdot2\sec\theta} \;=\;\tfrac{1}{4}\int \frac{\sec\theta}{\tan^2\theta}\,d\theta$ . . . etc.

Quote:

$2)\;\; e^{2x}\sin(e^x)\,dx$
$\begin{array}{ccccccc}u &=& e^x & & dv &=&e^x\sin(e^x)\,dx \\ du &=& e^x\,dx & & v &=& \text{-}\cos(e^x) \end{array}$

We have: . $\text{-}e^x\cos(e^x) + \int e^x\cos(e^x)\,dx \;=\;\text{-}e^x\cos(e^x) + \sin(e^x) + C$

Quote:

$3)\;\; \int \frac{dx}{\sqrt{x}(x^{\frac{1}{3}}+1)}$
Let $u = x^{\frac{1}{6}} \quad\Rightarrow\quad x = u^6 \quad\Rightarrow\quad dx = 6x^5\,du$
. . Also: . $x^{\frac{1}{2}} = u^3,\;\;x^{\frac{1}{3}} = u^2$

Substitute: . $\int\frac{6u^5\,du}{u^3(u^2+1)} \;=\;6\!\!\int\frac{u^2}{u^2+1}\,du$ . . . etc.

Quote:

$4)\;\;\int\frac{x^2+1}{x^3-4x^2}\,dx$
Partial Fractions: . $\frac{x^2+1}{x^2(x-4)} \;=\;\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-4}$

. . and we get: . $A = \text{-}\tfrac{1}{16},\;B = \text{-}\tfrac{1}{4},\;C = \tfrac{17}{16}$

So we have: . $\int\left[\frac{\text{-}\frac{1}{16}}{x} + \frac{\text{-}\frac{1}{4}}{x^2} + \frac{\frac{17}{16}}{x-4}\right]\,dx \;=\;\text{-}\tfrac{1}{16}\!\!\int\frac{dx}{x} - \tfrac{1}{4}\!\!\int x^{-2}dx + \tfrac{17}{16}\!\!\int\frac{dx}{x-4} \quad\hdots$

• October 31st 2008, 05:20 AM
tester85
For Qn 13 and 14. You can refer to the link below for step by step explanation and guide. Hope it helps.

Antiderivatives / Reduction Formulas - 2