Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{x^{2}-1}{x^{2}+4} d x$

Solution:

Add and subtract 4 in the numerator, we get

$=\int \frac{x^{2}+4-4-1}{x^{2}+4}=\int \frac{\left(x^{2}+4\right)-4-1}{x^{2}+4} d x$

$=\int \frac{\left(x^{2}+4\right)-5}{x^{2}+4} d x=\int \frac{\left(x^{2}+4\right)}{x^{2}+4} d x-\int \frac{5}{x^{2}+4} d x$ \{separate the numerator terms)

$=\int d x-\int \frac{5}{x^{2}+4} d x=\int d x-5 \int \frac{1}{x^{2}+4} d x$

$=\int \mathrm{dx}-5 \int \frac{1}{\mathrm{x}^{2}+2^{2}} \mathrm{dx}=\mathrm{x}-5 \times \frac{1}{2} \tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{c}\left\{\operatorname{since} \int \frac{1}{\mathrm{x}^{2}+\mathrm{a}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)+\mathrm{c}\right\}$

$=\mathrm{x}-\frac{5}{2} \tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{c}$

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