Integration Help: Solve in 3 Hours, Steps Included

[ineedhelpnow]

please help! this homework assignment is due in like 3 hours and i have to get it done.

$$\int \frac{1 \, dx}{(x^2+8x+17)^{2}}$$

$$\int_{-1/ \sqrt{3}}^{1/ \sqrt{3}} \frac{e^{arctan {y}} \, dy}{(1+y^2)}$$

i need to see all the steps.

do i use partial fractions for the first one?

 

[Ackbach]

Is this assignment for a grade?

 

[ineedhelpnow]

Yeah. It was already due tho

 

[MarkFL]

Yeah. It was already due tho

The reason Ackbach asked if these problems are part of a graded assignment is that it is against our policy to knowingly help with problems that contribute to a student's final grade, in the interest of academic honesty.

But, since the deadline has passed, let's look at these problems so you can solve them for your own knowledge.

For the first problem, I suggest completing the square on the quadratic being squared in the denominator of the integrand, and then making an appropriate trigonometric substitution. What form do you get?

 

[Prove It]

please help! this homework assignment is due in like 3 hours and i have to get it done.

$$\int \frac{1 \, dx}{(x^2+8x+17)^{2}}$$

$$\int_{-1/ \sqrt{3}}^{1/ \sqrt{3}} \frac{e^{arctan {y}} \, dy}{(1+y^2)}$$

i need to see all the steps.

do i use partial fractions for the first one?

In the second, you should know that $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ \arctan{(x)} \right] = \frac{1}{1 + x^2} \end{align*}$, so a substitution of the form $\displaystyle \begin{align*} u = \arctan{(y)} \implies \mathrm{d}u = \frac{1}{1 + y^2} \, \mathrm{d}y \end{align*}$ is appropriate. Also note that $\displaystyle \begin{align*} u \left( -\frac{1}{\sqrt{3}} \right) = -\frac{\pi}{6} \end{align*}$ and $\displaystyle \begin{align*} u \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}{\frac{\mathrm{e}^{\arctan{(y)}}}{1 + y^2}\,\mathrm{d}y} = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}{\mathrm{e}^u\,\mathrm{d}u} \end{align*}$

The rest should be easy...

 

[chisigma]

please help! this homework assignment is due in like 3 hours and i have to get it done.

$$\int \frac{dx}{(x^2+8x+17)^{2}}$$

i need to see all the steps.

do i use partial fractions?...

The integral isn't trivial and the best way is to learn a general way to solve integrals of the type...

$\displaystyle \int \frac{d x}{(a + b\ x + c\ x^{2})^{n}}\ (1)$

If we write $\displaystyle R= a + b\ x + c\ x^{2}$ and $\displaystyle \Delta = 4\ a\ c - b^{2}$, integrating by parts we arrive at the useful formula...

$\displaystyle \int \frac{d x}{R^{n+1}} = \frac{b + 2\ c\ x}{n\ \Delta\ R^{n}} + \frac{(4\ n -2)\ c}{n\ \Delta}\ \int \frac{d x}{R^{n}},\ n>0 \ (2)$

... and then for n=1 is...

$\displaystyle \int \frac{d x}{R} = \frac{-2}{\sqrt{- \Delta}}\ \tanh^{-1} \frac{b + 2\ c\ x}{\sqrt{-\Delta}}\ \text{if}\ \Delta<0$

$\displaystyle = - \frac{2}{b + 2\ c\ x}\ \text{if}\ \Delta=0$

$\displaystyle = \frac{2}{\sqrt{\Delta}}\ \tan^{-1} \frac{b + 2\ c\ x}{\sqrt{\Delta}}\ \text{if}\ \Delta>0\ (3)$

Kind regards

$\chi$ $\sigma$

 

[ineedhelpnow]

isnt $\int \ \frac{4}{6u+4}du=\frac{2}{3}\ln\left({6u+4}\right)$

my calculator keeps giving me $\int \ \frac{4}{6u+4}du=\frac{2}{3}\ln\left({3u+2}\right)$

 

[Prove It]

isnt $\int \ \frac{4}{6u+4}du=\frac{2}{3}\ln\left({6u+4}\right)$

my calculator keeps giving me $\int \ \frac{4}{6u+4}du=\frac{2}{3}\ln\left({3u+2}\right)$

Well actually $\displaystyle \begin{align*} \int{ \frac{4}{6u+4}\,\mathrm{d}u} = \frac{2}{3} \ln{ \left| 6u + 4 \right| } + C \end{align*}$.

Now notice that

$\displaystyle \begin{align*} \ln{ \left| 6u + 4 \right| } &= \ln{ \left( 2 \left| 3u + 2 \right| \right) } \\ &= \ln{(2)}+ \ln{ \left| 3u + 2 \right| } \end{align*}$

So the two answers only vary by a constant, and since the integration constant is arbitrary anyway, that means that the two answers are equivalent AS LONG AS YOU WRITE + C AT THE END!