How to integrate $\int \frac{1}{1 + {(x + 2)}^{2}} d x$ $int 1/(1+(x+2)^2)dx$ ?

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Answer 1 · 2018-06-23T10:11:03

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let $u = x + 2$ $u = x+2$

$d u = d x$ $du = dx$

$\Rightarrow \int \frac{1}{1 + u^{2}} d u$ $=> int 1/(1+u^2) du$

$u = tan θ$ $u = tan theta$

$d u = {sec}^{2} θ d θ$ $du = sec^2 theta d theta$

$\Rightarrow \int \frac{{sec}^{2} θ d θ}{1 + {tan}^{2} θ}$ $=> int (sec^2 theta d theta ) /(1+tan^2 theta )$

$\Rightarrow \int \frac{{sec}^{2} θ d θ}{{sec}^{2} θ}$ $=> int (sec^2 theta d theta ) / sec^2 theta$

$\Rightarrow \int 1 d θ$ $=> int 1 d theta$

$\Rightarrow θ + c$ $=> theta + c$

$\Rightarrow arctan u + c$ $=> arctan u + c$

$= arctan (x + 2) + c$ $=color(red)( arctan(x+2) + c$

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