ln(x),cos(x),sin(x),ex.

pickslides · #2 $Old$ June 2nd, 2009, 04:15 PM

$\int e^x\sqrt{e^x-1} dx$

make $u = e^x-1 \Rightarrow \frac{du}{dx} = e^x$

giving $\int \frac{du}{dx}\sqrt{u} dx$

= $\int \sqrt{u} du$

= $\int u^{\frac{1}{2}} du$

= $\frac{2}{3}u^{\frac{3}{2}}+c$

= $\frac{2}{3}(e^x-1)^{\frac{3}{2}} +c$

Now sub in terminals and subtract.

putnam120 · #3 $Old$ June 2nd, 2009, 04:18 PM

All of these seem to be solvable using substitution.

1) Try using $u=\tan\left(\frac{x}{2}\right)$

2) $u=e^x-1$

3) $u=3x^2+2x-1$

Soroban · #4 $Old$ June 2nd, 2009, 09:17 PM

Hello, dhiab!

The third one is quite messy . . .

Quote:

$\int^3_1 x\ln(3x^2+2x-1)\,dx$

By parts: . $\begin{array}{ccccccc}u &=& \ln(3x^2+2x-1) & & dv&=& x\,dx \\ \\[-3mm] du &=& \dfrac{(6x+2)\,dx}{3x^2+2x-1} & & v &=& \frac{1}{2}x^2 \end{array}$

We have: . $\tfrac{1}{2}x^2\ln(3x^2+2x-1) - \int\tfrac{1}{2}x^2\,\frac{6x+2}{3x^2+2x-1}\,dx$

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

. . . . . . $= \;\tfrac{1}{2}x^2\ln(3x^2+2x-1) - \int\left[x -\tfrac{1}{3} + \frac{\frac{5}{3}x - \frac{1}{3}}{3x^2+2x-1}\right]\,dx$ . (long division)

. . . . . . $= \;\tfrac{1}{2}x^2\ln(3x^2+2x-1) - \int\left(x - \tfrac{1}{3}\right)\,dx - \tfrac{1}{3}\int\underbrace{\left[\frac{5x-1}{(x+1)(3x-1)}\right]}_{\text{partial fractions}}\,dx$

Good luck!

dhiab · #5 $Old$ June 3rd, 2009, 01:01 AM

thank you for all

Prove It · #6 $Old$ June 3rd, 2009, 06:37 AM

Quote:

Originally Posted by putnam120 $View Post$

All of these seem to be solvable using substitution.

1) Try using $u=\tan\left(\frac{x}{2}\right)$

2) $u=e^x-1$

3) $u=3x^2+2x-1$

The third one can not be solved using a u-substitution because $\frac{du}{dx} = 6x + 2$ , NOT x.

putnam120 · #7 $Old$ June 3rd, 2009, 09:43 AM

Quote:

Originally Posted by Prove It $View Post$

The third one can not be solved using a u-substitution because $\frac{du}{dx} = 6x + 2$ , NOT x.

Yes I am aware of that, but you can note that $\int_1^3 x\ln(3x^2+2x-1)dx=\frac{1}{6}\int_1^3 6x\ln(3x^2+2x-1)dx$ . Now you should be able to solve $2\int_1^3\ln(3x^2+2x-1)dx$ with integration by parts.

Putting this all together it would be "easier" instead solve $\int_1^3(6x+2)\ln(3x^2+2x-1)dx$ and then do the necessary algebraic manipulations to find the value of $\int_1^3 x\ln(33x^2+2x-1)dx$ .

As for the hint to number (1) things will work out as follows:

$dx=\frac{2du}{1+u^2}$

$\cos(x)=\frac{1-u^2}{1+u^2}$

$\sin(x)=\frac{2u}{1+u^2}$

You can verify these by drawing the necessary right triangle.