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

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

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

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

= $\displaystyle \int \sqrt{u} du$

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

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

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

Now sub in terminals and subtract.

3. All of these seem to be solvable using substitution.

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

2) $\displaystyle u=e^x-1$

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

4. Hello, dhiab!

The third one is quite messy . . .

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

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

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

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

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

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

Good luck!

5. thank you for all

6. Originally Posted by putnam120
All of these seem to be solvable using substitution.

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

2) $\displaystyle u=e^x-1$

3) $\displaystyle u=3x^2+2x-1$
The third one can not be solved using a u-substitution because $\displaystyle \frac{du}{dx} = 6x + 2$, NOT x.

7. Originally Posted by Prove It
The third one can not be solved using a u-substitution because $\displaystyle \frac{du}{dx} = 6x + 2$, NOT x.
Yes I am aware of that, but you can note that $\displaystyle \int_1^3 x\ln(3x^2+2x-1)dx=\frac{1}{6}\int_1^3 6x\ln(3x^2+2x-1)dx$. Now you should be able to solve $\displaystyle 2\int_1^3\ln(3x^2+2x-1)dx$ with integration by parts.

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

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

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

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

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

You can verify these by drawing the necessary right triangle.