int (1 + 2x + 3x^2 + 4x^3 + dots) , dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given series is $1 + 2x + 3x^2 + 4x^3 + \dots = \sum_{n=1}^{\infty} n x^{n-1}$.This is the derivative of the geometric series $\sum_{n=0}^{\in

Vedclass · Accepted Answer

$(1 - x)^{-1} + c$

Vedclass · Answer

(B) The given series is $1 + 2x + 3x^2 + 4x^3 + \dots = \sum_{n=1}^{\infty} n x^{n-1}$.This is the derivative of the geometric series $\sum_{n=0}^{\infty} x^n = (1 - x)^{-1}$ for $|x| Thus,$1 + 2x + 3x^2 + 4x^3 + \dots = \frac{d}{dx} (1 - x)^{-1} = (1 - x)^{-2}$.Now,we integrate this expression:$\int (1 - x)^{-2} \, dx = \frac{(1 - x)^{-1}}{-1 \cdot (-1)} + c = (1 - x)^{-1} + c$.

$\int (1 + 2x + 3x^2 + 4x^3 + \dots) \, dx = $

Similar Questions

Integrate the function $\tan ^{2}(2 x-3)$.

Find the following integral:
$\int \frac{1-\sin x}{\cos ^{2} x} d x$

$\int \csc^4 x \, dx = $

The value of $\int \frac{d x}{7+6 x-x^2}$ is equal to

$\int (\sec x + \tan x)^2 dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module