How do you integrate $\int \frac{d x}{{(4 x^{2} + 9)}^{2}}$ $int dx/(4x^2+9)^2$ using trig substitutions?

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How do you integrate $\int \frac{d x}{{(4 x^{2} + 9)}^{2}}$ $int dx/(4x^2+9)^2$ using trig substitutions?

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Answer 1 · 2018-03-21T02:16:02

$\int \frac{d x}{{(4 x^{2} + 9)}^{2}} = \frac{1}{108} arctan (\frac{2 x}{3}) + \frac{x}{72 x^{2} + 162} + C$ $int dx/(4x^2+9)^2=1/108arctan((2x)/3)+x/(72x^2+162)+C$

Explanation:

$\int \frac{d x}{{(4 x^{2} + 9)}^{2}}$ $int dx/(4x^2+9)^2$

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

After using $2 x = 3 tan u$ $2x=3tanu$ and $2 d x = 3 {(sec u)}^{2} \cdot d u$ $2dx=3(secu)^2*du$ transforms, this integral became

$\frac{1}{2} \int \frac{3 {(sec u)}^{2} \cdot d u}{81 {(sec u)}^{4}}$ $1/2int (3(secu)^2*du)/(81(secu)^4)$

= $\frac{1}{54} \int {(cos u)}^{2} \cdot d u$ $1/54int (cosu)^2*du$

= $\frac{1}{108} \int (1 + cos 2 u) \cdot d u$ $1/108int (1+cos2u)*du$

= $\frac{1}{108} u + \frac{1}{216} sin 2 u + C$ $1/108u+1/216sin2u+C$

= $\frac{1}{108} u + \frac{1}{216} \cdot \frac{2 tan u}{{(tan u)}^{2} + 1} + C$ $1/108u+1/216*(2tanu)/((tanu)^2+1)+C$

After using $2 x = 3 tan u$ $2x=3tanu$ , $tan u = \frac{2 x}{3}$ $tanu=(2x)/3$ and $u = arctan (\frac{2 x}{3})$ $u=arctan((2x)/3)$ inverse transforms, I found

$\int \frac{d x}{{(4 x^{2} + 9)}^{2}}$ $int dx/(4x^2+9)^2$

= $\frac{1}{108} arctan (\frac{2 x}{3}) + \frac{1}{216} \cdot \frac{2 \cdot \frac{2 x}{3}}{{(\frac{2 x}{3})}^{2} + 1} + C$ $1/108arctan((2x)/3)+1/216*(2*(2x)/3)/(((2x)/3)^2+1)+C$

= $\frac{1}{108} arctan (\frac{2 x}{3}) + \frac{x}{72 x^{2} + 162} + C$ $1/108arctan((2x)/3)+x/(72x^2+162)+C$

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How do you integrate $\int \frac{d x}{{(4 x^{2} + 9)}^{2}}$ $int dx/(4x^2+9)^2$ using trig substitutions?

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Explanation:

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How do you integrate $\int \frac{d x}{{(4 x^{2} + 9)}^{2}}$ $int dx/(4x^2+9)^2$ using trig substitutions?