How do I find the integral $\int \frac{x^{3} + 4}{x^{2} + 4} d x$ $int(x^3+4)/(x^2+4)dx$ ?

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How do I find the integral $\int \frac{x^{3} + 4}{x^{2} + 4} d x$ $int(x^3+4)/(x^2+4)dx$ ?

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Answer 1 · 2014-08-01T20:02:15

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

Explanation,

$I = \int \frac{x^{3} + 4}{x^{2} + 4} d x$ $I=int(x^3+4)/(x^2+4)dx$

$I = \int (\frac{x^{3}}{x^{2} + 4} + \frac{4}{x^{2} + 4}) d x$ $I=int((x^3)/(x^2+4)+4/(x^2+4))dx$

$I = \int \frac{x^{3}}{x^{2} + 4} d x + \int \frac{4}{x^{2} + 4} d x$ $I=int(x^3)/(x^2+4)dx+int4/(x^2+4)dx$

$I = I_{1} + I_{2}$ $I=I_1+I_2$ ....... $(i)$ $(i)$

Now considering only first integral, which is $I_{1} = \int \frac{x^{3}}{x^{2} + 4} d x$ $I_1=int(x^3)/(x^2+4)dx$

Using Integration by Substitution,

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

$= \int \frac{t}{2} \cdot \frac{1}{t + 4} \cdot d t$ $=intt/2*1/(t+4)*dt$

$= \frac{1}{2} \int \frac{t}{t + 4} \cdot d t$ $=1/2intt/(t+4)*dt$ , this can be written as

$= \frac{1}{2} \int \frac{t + 4 - 4}{t + 4} \cdot d t$ $=1/2int(t+4-4)/(t+4)*dt$

$= \frac{1}{2} \int (1 - \frac{4}{t + 4}) d t$ $=1/2int(1-4/(t+4))dt$

$= \frac{1}{2} \int d t - 2 \int \frac{1}{t + 4} d t$ $=1/2intdt-2int1/(t+4)dt$

$= \frac{1}{2} t - 2 ln (t + 4) + c_{1}$ $=1/2t-2ln(t+4)+c_1$ , where $c_{1}$ $c_1$ is a constant

replacing $t$ $t$ , we get,

$I_{1} = \frac{1}{2} x^{2} - 2 ln (x^{2} + 4) + c_{1}$ $I_1=1/2x^2-2ln(x^2+4)+c_1$ , where $c_{1}$ $c_1$ is a constant

considering second integral

$I_{2} = \int \frac{4}{x^{2} + 4} d x$ $I_2=int4/(x^2+4)dx$

using Trigonometric Substitution to solve this problem,

$I_{2} = 4 \cdot \frac{1}{2} {tan}^{- 1} (\frac{x}{2}) + c_{2}$ $I_2=4*1/2tan^-1(x/2)+c_2$

$I_{2} = 2 {tan}^{- 1} (\frac{x}{2}) + c_{2}$ $I_2=2tan^-1(x/2)+c_2$

Finally, plugging in both $I_{1}$ $I_1$ and $I_{2}$ $I_2$ in $(i)$ $(i)$

$I = \frac{1}{2} x^{2} - 2 ln (x^{2} + 4) + c_{1} + 2 {tan}^{- 1} (\frac{x}{2}) + c_{2}$ $I=1/2x^2-2ln(x^2+4)+c_1+2tan^-1(x/2)+c_2$

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

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How do I find the integral $\int \frac{x^{3} + 4}{x^{2} + 4} d x$ $int(x^3+4)/(x^2+4)dx$ ?

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