using integration by parts find out the value of this ,i go by this way then i stucked>>>>>>
[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 $$\dfrac{2^{x+1}}{\ln(2)}$$
Please learn to use the new TeX \$\dfrac{2^{x+1}}{\ln(2)}\$
OR \$\$\dfrac{2^{x+1}}{\ln(2)}\$\$ gives $$\dfrac{2^{x+1}}{\ln(2)}$$
Please, please take a look at this. Also look at this.
Your answer is $\displaystyle\left. {{{{2^{x + 1}}} \over {\ln (2)}}} \right|_{ - 1}^1$
There is no need for parts in this question.
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>>>>
[ Okay, knowing that the derivative of $\displaystyle \frac{2^x}{ln 2}$ is $\displaystyle 2^x$, it follows immediately that $\displaystyle \int 2^x dx= \frac{2^x}{ln(2)}+ C$
I would first use a substitution. Since I now know how to integrate [latex]2^x[/latex] I would get back to that by letting $\displaystyle y= \sqrt{2x+ 1}= (2x+ 1)^{1/2}$. Then [tex]dy= (1/2)(2x+ 1)^{-1/2}(2)dx= \frac{dx}{\sqrt{2x+ 1}= \frac{dx}{y}[/latex] or [latex]ydy= dx[/latex],After that Using the parts$$\int_{-1}^{1}2^{\sqrt{2x+1}}$$
In this problem i must use parts>>>>
So in term of the new variable $\displaystyle y= \sqrt{2x+ 1}$ the integral is $\displaystyle \int y2^y dy$ and you can use "integration" by parts, taking [latex]u= y[/latex] and [latex]dv= 2^y[/latex]