# Math Help - Integral

1. ## Integral

using integration by parts find out the value of this ,i go by this way then i stucked>>>>>>

2. ## Re: Integral

Originally Posted by srirahulan
using integration by parts find out the value of this ,
Why use parts? What is the derivative of $\dfrac{2^{x+1}}{\ln(2)}~?$

3. ## Re: Integral

[QUOTE=srirahulan;810859]using integration by parts find out the value of this ,[/QUOTE

Please learn to use the new TeX \$\dfrac{2^{x+1}}{\ln(2)}\$

OR \$\$\dfrac{2^{x+1}}{\ln(2)}\$\$ gives $$\dfrac{2^{x+1}}{\ln(2)}$$

4. ## Re: Integral

Originally Posted by srirahulan
using integration by parts find out the value of this ,i go by this way then i stucked>>>>>>
Here i forgot to put sqrt in there>>>>

5. ## Re: Integral

Originally Posted by srirahulan
Here i forgot to put sqrt in there>>>>
You did not answer the question I posted. Why?

6. ## Re: Integral

So How can i do that easily??? ((please note i have correction in my problem))

7. ## Re: Integral

Originally Posted by srirahulan
So How can i do that easily??? ((please note i have correction in my problem))
Please learn to use the new TeX \$\dfrac{2^{x+1}}{\ln(2)}\$

OR \$\$\dfrac{2^{x+1}}{\ln(2)}\$\$ gives $$\dfrac{2^{x+1}}{\ln(2)}$$

Your answer is $\displaystyle\left. {{{{2^{x + 1}}} \over {\ln (2)}}} \right|_{ - 1}^1$

There is no need for parts in this question.

8. ## Re: Integral

Thank You Plato>>

I have problem like this,
First they ask me to proof this $$\frac{d\frac{2^{x}}{ln2}}{dx}=2^{x}$$,Then they ask to find out $$\int2^{x}$$,After that Using the parts$$\int_{-1}^{1}2^{\sqrt{2x+1}}$$

In this problem i must use parts>>>>

9. ## Re: Integral

Originally Posted by srirahulan
Thank You Plato>>

I have problem like this,
First they ask me to proof this $$\frac{d\frac{2^{x}}{ln2}}{dx}=2^{x}$$,Then they ask to find out $$\int2^{x}$$,After that Using the parts$$\int_{-1}^{1}2^{\sqrt{2x+1}}$$

In this problem i must use parts>>>>
Look at this.

10. ## Re: Integral

Explain How you use parts?? please explain

11. ## Re: Integral

Originally Posted by srirahulan
Explain How you use parts?? please explain

12. ## Re: Integral

[
Originally Posted by srirahulan
Thank You Plato>>

I have problem like this,
First they ask me to proof this $$\frac{d\frac{2^{x}}{ln2}}{dx}=2^{x}$$,Then they ask to find out $$\int2^{x}$$
Okay, knowing that the derivative of $\frac{2^x}{ln 2}$ is $2^x$, it follows immediately that $\int 2^x dx= \frac{2^x}{ln(2)}+ C$

,After that Using the parts$$\int_{-1}^{1}2^{\sqrt{2x+1}}$$

In this problem i must use parts>>>>
I would first use a substitution. Since I now know how to integrate $2^x$ I would get back to that by letting $y= \sqrt{2x+ 1}= (2x+ 1)^{1/2}$. Then [tex]dy= (1/2)(2x+ 1)^{-1/2}(2)dx= \frac{dx}{\sqrt{2x+ 1}= \frac{dx}{y}[/latex] or $ydy= dx$

So in term of the new variable $y= \sqrt{2x+ 1}$ the integral is $\int y2^y dy$ and you can use "integration" by parts, taking $u= y$ and $dv= 2^y$