Finding a certain antiderivative

Question

The problem says:

If $$f(x)=\frac{b_2}{x^2}+\frac{b_3}{x^3}+...+\frac{b_m}{x^m}$$ find a function $g$ with $g' = f$.

This problem is stated in the differentiation part of the book, integration comes later.

I tried starting with a simple example and trying to find a pattern from there:

If

$$f(x)=\frac{b_2}{x^2}$$

Then we can try:

$$g(x)=-\frac{b_2}{x}$$ $$g'(x)=-b_2\left(\frac{-1}{x^2}\right) = \frac{b_2}{x^2} = f(x)$$

Then if

$$f(x)=\frac{b_2}{x^2} + \frac{b_3}{x^3}$$

I tried:

$$g(x)=-\frac{b_2}{x}-\frac{b_3}{x^{3/2}}$$ $$g'(x)=-b_2\left(\frac{-1}{x^2}\right) - b_3 \left( \frac{-\frac{3}{2}x^{1/2}}{x^3} \right) $$

Which is not what I wanted. The denominator in the second summand is ok, but the numerator is not what I need.

But I'm stuck on what I need to look for, any hints for deriving the correct pattern would be appreciated.

Answer 1

$$\displaystyle f(x)=\frac{b_2}{x^2}+\frac{b_3}{x^3}+...+\frac{b_m}{x^m}$$

Say $g(x)=a_1(x)+a_2(x)+...+a_{m-1}(x)$, such that $a_i'(x)$ gives the $i^{th}$ term of $f(x)$; that is,

$$\displaystyle a_i'(x)=\frac{b_{i+1}}{x^{i+1}}$$

The first term of $\displaystyle f(x)$ has an $x^2$ in the denominator. You know that when we differentiate $1/x$, we get $x^2$ in the denominator. So $a_1(x)=a/x,a_1'(x)=-a/x^2=b_2/x^2\implies a=-b_2$.

The second term of $f(x)$ has an $x^3$ in the denominator. This is one degree higher than $x^2$, so it makes sense to chose $a_2(x)$ having $x^2$ in the denominator, which is one degree higher than $x$. $a_2(x)=c/x^2,a_2'(x)=-2c/x^3=b_3/x^3\implies c=-b_3/2$.

Similarly chose $a_3(x)$ having $x^3$ in the denominator, to obtain $\displaystyle a_3(x)=-\frac{b_4}{3x^3}$.

You might be able to see the pattern: $\displaystyle a_i(x)=-\frac{b_{i+1}}{ix^i}$

$$\therefore\displaystyle g(x)=-\frac{b_2}{x}-\frac{b_3}{2x^2}-\frac{b_4}{3x^3}...-\frac{b_m}{(m-1)x^{m-1}}+c$$

We put an arbitrary constant $c$ because it vanishes on differentiation.

Answer 2

Hint:

For any $\;2\le k\le m\;$ :

$$\int\left(\frac{b_k}{x^k}\right)dx=\frac{\,b_k}{(1-k)x^{k-1}}+C(=\text{constant})$$

The basic trick is

$$\forall\,-1\neq n\in\Bbb R\;,\;\;\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$$

Answer 3

$$f(x)=\sum_{i=2}^m\frac{b_i}{x^i}$$ $$\Longrightarrow g(x)=\int f(x)dx$$ $$=\int \sum_{i=2}^m\frac{b_i}{x^i}dx$$ $$=\sum_{i=2}^m \int\frac{b_i}{x^i}dx$$ $$=-\sum_{i=2}^m\frac{b_i}{(i-1)x^{i-1}}+C$$ which is the required function ($\because \int x^ndx=\frac{x^{n+1}}{n+1}+C$)

Hope it helps

Answer 4

Solve the problem for $$ p(x)=\frac{b}{x^k}=bx^{-k} $$ Then apply the trick to each summand and sum the functions you get. Since the derivative of a sum is the sum of the derivatives, you're done.

Now, if we want to find a function $q(x)$ such that $q'=p$, we may first try with a monomial, $q(x)=ax^h$. Then we know that $$ q'(x)=hax^{h-1} $$ so we can take $ha=b$ and $h-1=-k$. This means $h=-k+1$ and $$ a=\frac{b}{-k+1} $$ which is possible provided $k\ne1$.

Putting together the pieces, you can take $$ g(x)=\frac{b_2}{(-2+1)x}+\frac{b_3}{(-3+1)x^2}+\dots+\frac{b_m}{(-m+1)x^{m-1}} =-\frac{b_2}{x}-\frac{b_3}{2x^2}-\dots-\frac{b_m}{(m-1)x^{m-1}} $$