how do I find the antiderivative to

1 / (x(1+4x^2))

and

1/xsqrt(x+1)

?

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- Apr 8th 2008, 08:20 AMweasley74Antiderivative
how do I find the antiderivative to

1 / (x(1+4x^2))

and

1/xsqrt(x+1)

? - Apr 8th 2008, 11:37 AMabender
1 / (x(1+4x^2)) the same as ( 4x^3 + x )^-1 I am in class right now (so cannot go into too much depth here) , and you may know this already, but this problem will involve the natural logarithm (ex. integral 1/x = ln(x) )

You should get something*like***NOT EXACTLY**this:

A ln(x) + B ln( Cx^2 term + D) where A,B,C are constants (that you will fill out!)

Note that any plus could essentially be a minus (by virtue of + negNumber)

and that A B C or D could be 0 (or pos or neg).

Determine which method is the most appropriate: integration by parts, "normal" substitution, trig substitution (I am not ruling anything out, you can get this), integration by parts, inverse chain rule, the fundamental Thm of Calc, etc. etc. It may be even simpler.

Scroll below if you want to see my NON-worked out solution in case you simply want to check your answer when you are confident:

ln(x) - (1/2 (ln(4x^2 + 1))) and no constant in your solution since you are only asked for the antideriv, which means indefinite integral form. - Apr 8th 2008, 11:39 AMMoo
Hello,

Decompose it into partial fractions (i don't know the exact denomination, i think it's it : Partial fraction decomposition over the reals - Wikipedia, the free encyclopedia)

The principle is simple, the application is a bit less simple, following the case.

$\displaystyle \frac{1}{x(1+4x^2)}$

This consists in finding a, b and c such as :

$\displaystyle \frac{1}{x(1+4x^2)}=\frac{a}{x}+\frac{bx+c}{4x^2+1 }$

Do it by identification, while gathering it into a sole fraction :

$\displaystyle \frac{a}{x}+\frac{bx+c}{4x^2+1}=\frac{a(4x^2+1)}{x (4x^2+1)}+\frac{(bx+c)x}{x(4x^2+1)}$

$\displaystyle \frac{4ax^2+a+bx^2+cx}{x(4x^2+1)}$

$\displaystyle 4ax^2+a+bx^2+cx=(4a+b)x^2+cx+a=1=0x^2+0x+1$

So we have simultaneously :

$\displaystyle a=1$

$\displaystyle c=0$

$\displaystyle 4a+b=0 \rightarrow 4+b=0 \rightarrow b=-4$

Hence :

$\displaystyle \int \frac{1}{x(1+4x^2)} dx=\int \frac{1}{x}+\frac{-4x}{4x^2+1} dx$

$\displaystyle =\underbrace{\int \frac{1}{x} dx}_{\ln(x)+C} + \int \frac{-4x}{4x^2+1} dx$

the derivative of [tex]4x^2+1[tex] is 8x. So it'll be nice to make it appear at the numerator :

$\displaystyle \frac{-4x}{4x^2+1}=\frac{-1}{2} \frac{8x}{4x^2+1}=\frac{-1}{2} \frac{u'(x)}{u(x)}$

An antiderivate is thus $\displaystyle \frac{-1}{2} \ln(4x^2+1)$

So an antiderivate of $\displaystyle \frac{1}{x(1+4x^2)}$ is $\displaystyle \ln(x)-\frac{1}{2} \ln(4x^2+1)$

Phew - Apr 8th 2008, 11:56 AMabender
Yes, Moo's explanation/solution is probably more along the lines of what your teacher wants. It is also neater. I have yet to familiarize myself with this site's "MathType-ish" options. I did not even think of partial fractions. Thus, another way to do this (almost always in math...perhaps always). I love and don't love the fact that I tend to solve problems in the most unconventional ways. (good and sometimes bad thing).

- Apr 8th 2008, 11:58 AMMoo
Hi,

Partial fractions may be sort of a reflex when dealing with antiderivative. When it remains fractions with polynomials of degree less or equal to 3, it's ok. For 4, only a few of them. Any polynomials can be decomposed into polynomials of degree 1 and 2 (irreductible) in $\displaystyle \mathbb{R}$. In $\displaystyle \mathbb{C}$ any polynomial can be divided into polynomials of degree 1.

This technique is useful because you can quickly have the antiderivative when you know the derivatives of ln(u(x)) and arctan(u(x)) ;)

Quote:

You should get something like NOT EXACTLY this:

A ln(x) + B ln( Cx^2 term + D) where A,B,C are constants (that you will fill out!)