How do I find the integral $int(x^3+4)/(x^2+4)dx$ ?

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How do I find the integral $int(x^3+4)/(x^2+4)dx$ ?

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Answer 1 · 2014-08-01T20:02:15

$I=1/2x^2-2ln(x^2+4)+2tan^-1(x/2)+c$ , where $c$ is a constant

Explanation,

$I=int(x^3+4)/(x^2+4)dx$

$I=int((x^3)/(x^2+4)+4/(x^2+4))dx$

$I=int(x^3)/(x^2+4)dx+int4/(x^2+4)dx$

$I=I_1+I_2$ ....... $(i)$

Now considering only first integral, which is $I_1=int(x^3)/(x^2+4)dx$

Using Integration by Substitution,

let's $x^2=t$ , then $2xdx=dt$ yields, first integral

$=intt/2*1/(t+4)*dt$

$=1/2intt/(t+4)*dt$ , this can be written as

$=1/2int(t+4-4)/(t+4)*dt$

$=1/2int(1-4/(t+4))dt$

$=1/2intdt-2int1/(t+4)dt$

$=1/2t-2ln(t+4)+c_1$ , where $c_1$ is a constant

replacing $t$ , we get,

$I_1=1/2x^2-2ln(x^2+4)+c_1$ , where $c_1$ is a constant

considering second integral

$I_2=int4/(x^2+4)dx$

using Trigonometric Substitution to solve this problem,

$I_2=4*1/2tan^-1(x/2)+c_2$

$I_2=2tan^-1(x/2)+c_2$

Finally, plugging in both $I_1$ and $I_2$ in $(i)$

$I=1/2x^2-2ln(x^2+4)+c_1+2tan^-1(x/2)+c_2$

$I=1/2x^2-2ln(x^2+4)+2tan^-1(x/2)+c$ , where $c=c_1+c_2$ is again a constant

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How do I find the integral $int(x^3+4)/(x^2+4)dx$ ?

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