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

Aug 1, 2014

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

Explanation,

$I = \int \frac{{x}^{3} + 4}{{x}^{2} + 4} \mathrm{dx}$

$I = \int \left(\frac{{x}^{3}}{{x}^{2} + 4} + \frac{4}{{x}^{2} + 4}\right) \mathrm{dx}$

$I = \int \frac{{x}^{3}}{{x}^{2} + 4} \mathrm{dx} + \int \frac{4}{{x}^{2} + 4} \mathrm{dx}$

$I = {I}_{1} + {I}_{2}$ .......$\left(i\right)$

Now considering only first integral, which is ${I}_{1} = \int \frac{{x}^{3}}{{x}^{2} + 4} \mathrm{dx}$

let's ${x}^{2} = t$ , then $2 x \mathrm{dx} = \mathrm{dt}$ yields, first integral

$= \int \frac{t}{2} \cdot \frac{1}{t + 4} \cdot \mathrm{dt}$

$= \frac{1}{2} \int \frac{t}{t + 4} \cdot \mathrm{dt}$, this can be written as

$= \frac{1}{2} \int \frac{t + 4 - 4}{t + 4} \cdot \mathrm{dt}$

$= \frac{1}{2} \int \left(1 - \frac{4}{t + 4}\right) \mathrm{dt}$

$= \frac{1}{2} \int \mathrm{dt} - 2 \int \frac{1}{t + 4} \mathrm{dt}$

$= \frac{1}{2} t - 2 \ln \left(t + 4\right) + {c}_{1}$, where ${c}_{1}$ is a constant

replacing $t$, we get,

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

considering second integral

${I}_{2} = \int \frac{4}{{x}^{2} + 4} \mathrm{dx}$

using Trigonometric Substitution to solve this problem,

${I}_{2} = 4 \cdot \frac{1}{2} {\tan}^{-} 1 \left(\frac{x}{2}\right) + {c}_{2}$

${I}_{2} = 2 {\tan}^{-} 1 \left(\frac{x}{2}\right) + {c}_{2}$

Finally, plugging in both ${I}_{1}$ and ${I}_{2}$ in $\left(i\right)$

$I = \frac{1}{2} {x}^{2} - 2 \ln \left({x}^{2} + 4\right) + {c}_{1} + 2 {\tan}^{-} 1 \left(\frac{x}{2}\right) + {c}_{2}$

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