How do you integrate $\frac{1 + x}{1 + x^{2}} d x$ $(1+ x) / (1 + x^2) dx$ ?

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How do you integrate $\frac{1 + x}{1 + x^{2}} d x$ $(1+ x) / (1 + x^2) dx$ ?

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Answer 1 · 2016-06-09T23:02:53

$arctan (x) + \frac{1}{2} ln (1 + x^{2}) + C$ $arctan(x)+1/2ln(1+x^2)+C$

Explanation:

We have:

$\int \frac{1 + x}{1 + x^{2}} d x$ $int(1+x)/(1+x^2)dx$

Split up the numerator and into two different integrals:

$= \int \frac{1}{1 + x^{2}} d x + \int \frac{x}{1 + x^{2}} d x$ $=int1/(1+x^2)dx+intx/(1+x^2)dx$

Notice that the first derivative is just the derivative of the arctangent function, that is, $\int \frac{1}{1 + x^{2}} d x = arctan (x) + C$ $int1/(1+x^2)dx=arctan(x)+C$ .

$= arctan (x) + \int \frac{x}{1 + x^{2}} d x$ $=arctan(x)+intx/(1+x^2)dx$

For the remaining integral, let $u = 1 + x^{2}$ $u=1+x^2$ and $d u = 2 x d x$ $du=2xdx$ .

$= arctan (x) + \frac{1}{2} \int \frac{2 x}{1 + x^{2}} d x$ $=arctan(x)+1/2int(2x)/(1+x^2)dx$

$= arctan (x) + \frac{1}{2} \int \frac{d u}{u}$ $=arctan(x)+1/2int(du)/u$

This is the natural logarithm integral: $\int \frac{d u}{u} = ln (| u |) + C$ $int(du)/u=ln(absu)+C$

$= arctan (x) + \frac{1}{2} ln (| u |) + C$ $=arctan(x)+1/2ln(absu)+C$

Since $u = 1 + x^{2}$ $u=1+x^2$ :

$= arctan (x) + \frac{1}{2} ln (1 + x^{2}) + C$ $=arctan(x)+1/2ln(1+x^2)+C$

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How do you integrate $\frac{1 + x}{1 + x^{2}} d x$ $(1+ x) / (1 + x^2) dx$ ?

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