How do you evaluate the integral $\int \frac{d x}{x^{3} + x}$ $int dx/(x^3+x)$ ?

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How do you evaluate the integral $\int \frac{d x}{x^{3} + x}$ $int dx/(x^3+x)$ ?

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Answer 1 · 2017-01-08T02:58:02

$\int \frac{d x}{x^{3} + x} = \frac{1}{2} ln (\frac{x^{2}}{x^{2} + 1}) + C$ $int (dx)/(x^3+x) = 1/2 ln (x^2/(x^2+1))+C$

Explanation:

We can write the integral as:

$\int \frac{d x}{x^{3} + x} = \int \frac{d x}{x (1 + x^{2})}$ $int (dx)/(x^3+x) = int (dx)/(x(1+x^2))$

now we can substitute:

$x = tan t$ $x=tant$

$d x = \frac{d t}{{cos}^{2} t}$ $dx= dt/cos^2t$

so we have:

$\int \frac{d x}{x^{3} + x} = \int \frac{d t}{{cos}^{2} t tan t (1 + {tan}^{2} t)}$ $int (dx)/(x^3+x) = int (dt)/ (cos^2t tan t (1+tan^2t))$

and using the trigonometric identity:

$1 + {tan}^{2} t = 1 + \frac{{sin}^{2} t}{{cos}^{2} t} = \frac{{cos}^{2} t + {sin}^{2} t}{{cos}^{2} t} = \frac{1}{{cos}^{2} t}$ $1+tan^2t = 1+sin^2t/cos^2t = (cos^2t+sin^2t)/cos^2t = 1/cos^2t$

$\int \frac{d x}{x^{3} + x} = \int \frac{d t}{{cos}^{2} t tan t \frac{1}{{cos}^{2} t}} = \int \frac{d t}{tan t} = \int \frac{cos t d t}{sin t}$ $int (dx)/(x^3+x) = int (dt)/ (cos^2t tan t 1/cos^2t)=int (dt)/tant= int (costdt)/sint$

We can see that: $cos t d t = d (sin t)$ $cost dt = d(sint)$ , so:

$\int \frac{d x}{x^{3} + x} = \int \frac{d sin t}{sin t} = ln | sin t | + C$ $int (dx)/(x^3+x) = int (dsint)/sint = ln abs sin t +C$

To substitute back $x$ $x$ we can note that:

$x = tan t = \frac{sin t}{\sqrt{1 - {sin}^{2} t}}$ $x=tant = sint/sqrt(1-sin^2t)$

$x^{2} = \frac{{sin}^{2} t}{1 - {sin}^{2} t}$ $x^2= sin^2t/(1-sin^2t)$

$x^{2} (1 - {sin}^{2} t) = {sin}^{2} t$ $x^2(1-sin^2t)= sin^2t$

$x^{2} - x^{2} {sin}^{2} t = {sin}^{2} t$ $x^2-x^2sin^2t= sin^2t$

$x^{2} = x^{2} {sin}^{2} t + {sin}^{2} t$ $x^2=x^2sin^2t+ sin^2t$

$x^{2} = (x^{2} + 1) {sin}^{2} t$ $x^2=(x^2+1)sin^2t$

$\frac{x^{2}}{x^{2} + 1} = {sin}^{2} t$ $x^2/(x^2+1) = sin^2t$

$\sqrt{\frac{x^{2}}{x^{2} + 1}} = | sin t |$ $sqrt(x^2/(x^2+1)) = abs sint$

so that finally:

$\int \frac{d x}{x^{3} + x} = ln \sqrt{\frac{x^{2}}{x^{2} + 1}} + C = \frac{1}{2} ln (\frac{x^{2}}{x^{2} + 1}) + C$ $int (dx)/(x^3+x) = ln sqrt(x^2/(x^2+1)) +C = 1/2 ln (x^2/(x^2+1))+C$

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How do you evaluate the integral $\int \frac{d x}{x^{3} + x}$ $int dx/(x^3+x)$ ?

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