# How do you integrate #f(x) = 2x+4x^3+6x^5+...+2nx^(2n-1)+...# ?

#int f(x) dx = 1/(1-x^2) + C# for#x in (-1, 1)#

For finite sums we know that the integral of the sum is equal to the sum of the integrals.

This extends to infinite sums, provided we know that the sums converge.

The slight trick here is that the constant of integration can be adjusted to give us a familiar sum.

By signing up, you agree to our Terms of Service and Privacy Policy

To integrate the function ( f(x) = 2x + 4x^3 + 6x^5 + \ldots + 2nx^{2n-1} + \ldots ), we'll look at the general term of the series ( 2nx^{2n-1} ) to find its antiderivative. The antiderivative of ( x^n ) with respect to ( x ) is ( \frac{x^{n+1}}{n+1} ).

So, for the term ( 2nx^{2n-1} ), the antiderivative is ( \frac{2n}{2n}x^{2n} = x^{2n} ).

Therefore, the antiderivative of ( f(x) ) is:

[ F(x) = x^2 + x^4 + x^6 + \ldots + x^{2n} + \ldots + C ]

where ( C ) is the constant of integration.

So, the integral of ( f(x) ) is:

[ \int f(x) , dx = \int (2x + 4x^3 + 6x^5 + \ldots + 2nx^{2n-1} + \ldots) , dx = x^2 + x^4 + x^6 + \ldots + x^{2n} + \ldots + C ]

This is the indefinite integral of the given series ( f(x) ).

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7