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Taylor series Calculator

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1

Solved example of power series

$\int\left(\frac{\arctan\left(x\right)}{x}\right)dx$
2

Rewrite the function $\arctan\left(x\right)$ as it's representation in Maclaurin series expansion

$\int\frac{\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{2n+1}x^{\left(2n+1\right)}}{x}dx$
3

Bring the denominator $x$ inside the power serie

$\int\sum_{n=0}^{\infty } \frac{\frac{{\left(-1\right)}^n}{2n+1}x^{\left(2n+1\right)}}{x}dx$

Multiplying the fraction by $x^{\left(2n+1\right)}$

$\int\sum_{n=0}^{\infty } \frac{\frac{{\left(-1\right)}^nx^{\left(2n+1\right)}}{2n+1}}{x}dx$

Divide fractions $\frac{\frac{{\left(-1\right)}^nx^{\left(2n+1\right)}}{2n+1}}{x}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$

$\int\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{\left(2n+1\right)}}{\left(2n+1\right)x}dx$

Simplify the fraction $\frac{{\left(-1\right)}^nx^{\left(2n+1\right)}}{\left(2n+1\right)x}$ by $x$

$\int\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{2n}}{2n+1}dx$
4

Simplify the expression inside the integral

$\int\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{2n}}{2n+1}dx$
5

We can rewrite the power series as the following

$\sum_{n=0}^{\infty } \frac{1}{2n+1}\int{\left(-1\right)}^nx^{2n}dx$
6

The integral of a function times a constant (${\left(-1\right)}^n$) is equal to the constant times the integral of the function

$\sum_{n=0}^{\infty } \frac{1}{2n+1}{\left(-1\right)}^n\int x^{2n}dx$

Multiplying the fraction by ${\left(-1\right)}^n$

$\sum_{n=0}^{\infty } \frac{1{\left(-1\right)}^n}{2n+1}\int x^{2n}dx$

Any expression multiplied by $1$ is equal to itself

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{2n+1}\int x^{2n}dx$
7

Simplify the expression inside the integral

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{2n+1}\int x^{2n}dx$
8

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2n$

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{2n+1}\frac{x^{\left(2n+1\right)}}{2n+1}$
9

Multiplying fractions $\frac{{\left(-1\right)}^n}{2n+1} \times \frac{x^{\left(2n+1\right)}}{2n+1}$

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{\left(2n+1\right)}}{\left(2n+1\right)^2}$
10

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{\left(2n+1\right)}}{\left(2n+1\right)^2}+C_0$

Final Answer

$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{\left(2n+1\right)}}{\left(2n+1\right)^2}+C_0$

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