All!

Please, can someone help me with:

$\displaystyle \int\frac{x-3}{x^2+1}dx$

The result should be:

$\displaystyle \frac{1}{2}ln|x^2+1|-3arctanx+C$

Many thanks!

Results 1 to 8 of 8

- Jan 16th 2013, 07:05 AM #1

- Joined
- Jan 2013
- From
- Pizza
- Posts
- 57

- Jan 16th 2013, 07:17 AM #2
## Re: Int 1

Write this as

$\displaystyle \int\frac{x}{x^2+1}dx-3\int \frac{1}{1+x^2}dx$

For the first integral, substitute $\displaystyle u=1+x^2$.

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

$\displaystyle =\frac{1}{2}\ln |u|+\text{Constant} = \frac{1}{2}\ln |1+x^2| +\text{Constant}$

For the second integral, substitute $\displaystyle x=\tan \theta$.

$\displaystyle \int \frac{1}{1+x^2}dx = \int \frac{\sec^2 \theta}{1+\tan^2 \theta}d\theta$

Note that $\displaystyle 1+\tan^2 \theta =\sec^2 \theta$, so our integral will become

$\displaystyle \int d\theta = \theta + \text{Constant} = \arctan(x)+\text{Constant}$

Combine these two results to obtain

$\displaystyle \int\frac{x}{x^2+1}dx-3\int \frac{1}{1+x^2}dx$

$\displaystyle = \frac{1}{2}\ln|1+x^2|-3\arctan(x)+\text{Constant}$

- Jan 19th 2013, 09:22 AM #3

- Joined
- Jan 2013
- From
- Pizza
- Posts
- 57

- Jan 19th 2013, 12:57 PM #4

- Joined
- Apr 2005
- Posts
- 20,044
- Thanks
- 3173

## Re: Int 1

Have you never taken a Calculus class? What sbhatnagar did is pretty standard. To integrate $\displaystyle \int \frac{x}{x^2+ 1}dx$, seeing that "$\displaystyle x^2$" in the denominator and the "xdx" in the numerator, you should immediately think to try the substitution $\displaystyle u= x^2+ 1$ so that $\displaystyle du= 2xdx$ or $\displaystyle \frac{1}{2}du= x dx$. We can write

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

And now you should recognise that $\displaystyle \frac{d ln(u)}{du}= \frac{1}{u}$ so that $\displaystyle \int \frac{du}{u}= ln|u|+ C$.

(Note the absolute value: ln|u|. ln(u) is only defined for postive u and u, here, might be negative.)

- Jan 20th 2013, 07:07 AM #5

- Joined
- Jan 2013
- From
- Pizza
- Posts
- 57

- Jan 20th 2013, 07:10 AM #6

- Joined
- Jan 2013
- From
- Pizza
- Posts
- 57

- Jan 20th 2013, 01:16 PM #7

- Joined
- Oct 2008
- Posts
- 1,035
- Thanks
- 49

## Re: Int 1

Just in case a picture helps...

That's how the substitution works. However, you can see that it depends on the chain rule, and I would prefer to see how we are working backwards through a differentiation according to the chain rule.

If, like me, you were taught to see differentiation as a downwards journey and integration up, then you'll see the logic of this...

... where (key in spoiler) ...

__Spoiler__:

So, the bottom row shows that we have multiplied the derivative**with respect to the inner function**by the derivative of the inner function itself. Which is what the chain rule wants.

However, the inner function is just x, so the derivative of that is just one, so no big deal from applying the chain rule.

But now, carefully, remove the inner function (by twisting gently) and replace with tan theta...

Now the derivative of the inner function is sec squared so, as sbhatnagar explains, we are really integrating (or anti-differentiating) one, with respect to theta...

Then just replace the original inner function...

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

- Jan 20th 2013, 01:52 PM #8

- Joined
- Oct 2008
- Posts
- 1,035
- Thanks
- 49

## Re: Int 1

The first integral is of the other main kind (that involve a substitution). It comes ready with an appropriate inner function...

So you only need to deduce that the function F is ln...

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!