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Question: Evaluate the following integral: \(\int{(1\ +\ 2x\ +\ 3{{x}^{2}}\ +\ 4{{x}^{2}}\ +\ ........)\ dx}...

Evaluate the following integral:
∫(1 + 2x + 3x2 + 4x2 + ........) dx\int{(1\ +\ 2x\ +\ 3{{x}^{2}}\ +\ 4{{x}^{2}}\ +\ ........)\ dx}

Explanation

Solution

Hint: After integrating the function, identify the resulting progression (sequence) and consider it an infinite series, find the sum of Infinite series.

Complete step-by-step answer:
Let us consider the given integral ∫(1 + 2x + 3x2 + 4x2 + ........) dx\int{(1\ +\ 2x\ +\ 3{{x}^{2}}\ +\ 4{{x}^{2}}\ +\ ........)\ dx}. We know the integration formula, ∫x= xn+1n+1\int{x}=\ \dfrac{{{x}^{n+1}}}{n+1}. So, we will use it further as,
=∫(1 + 2x + 3x2 + 4x2 + ........) dx=\int{(1\ +\ 2x\ +\ 3{{x}^{2}}\ +\ 4{{x}^{2}}\ +\ ........)\ dx}
We know ∫(f(x)+g(x))dx= ∫f(x)dx+∫g(x)dx\int{\left( f\left( x \right)+g\left( x \right) \right)}dx=\ \int{f\left( x \right)dx+}\int{g}\left( x \right)dx. Now, integrating each term we get,
=x+2x22+3x33+4x44+......+C=x+2\dfrac{{{x}^{2}}}{2}+3\dfrac{{{x}^{3}}}{3}+4\dfrac{{{x}^{4}}}{4}+......+C
Cancelling the common terms in each term, we get
=x + x2 + x3 + ..................+C=x\ +\ {{x}^{2}}\ +\ {{x}^{3}}\ +\ ..................+C
Let's consider there are n terms. So, we can write it as,
=x + x2 + x3 + ................up to n terms+C=x\ +\ {{x}^{2}}\ +\ {{x}^{3}}\ +\ ................up\ to\ n\ terms+C
When we observe the sequence carefully, it forms a G.P. There is a common ratio between the terms. Considering each term, we get thet x2x = x3 x2 = x\dfrac{{{x}^{2}}}{x}\ =\ \dfrac{{{x}^{3}}\ }{{{x}^{2}}}\ =\ x.
We know that the first term of a GP = a and the common ratio = r. Observing the resulting sequence, we get that a=xa=x and r=xr=x.
We also know that the sum of an infinite G.P is equal to  a1−r\ \dfrac{a}{1-r}. Then we can write the sum of the resulted sequence =x + x2 + x3 + ................up to n terms=x\ +\ {{x}^{2}}\ +\ {{x}^{3}}\ +\ ................up\ to\ n\ terms as x1−x\dfrac{x}{1-x}.
Hence, we can write the integral as ∫(1 + 2x + 3x2 + 4x2 + ........) dx =x1−x+C\int{(1\ +\ 2x\ +\ 3{{x}^{2}}\ +\ 4{{x}^{2}}\ +\ ........)\ dx}\ =\dfrac{x}{1-x}+C.

Note: Identifying the G.P (Geometric Progression) series in an important step here. We cannot consider the sequence carrying infinite terms as infinite geometric series converges if its common ratio r satisfies −1<r<1{-1} < {r} < {1}. The mistake that can be made is by considering the sum of finite GP instead of infinite GP. This question is related to the evaluation of an indefinite integral, so we must not forget to add the constant term at the end.