How do you find the integral of $\int \frac{17}{16 + x^{2}} d x$ $int 17/(16+x^2)dx$ ?

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How do you find the integral of $\int \frac{17}{16 + x^{2}} d x$ $int 17/(16+x^2)dx$ ?

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Answer 1 · 2016-12-09T21:01:25

$\frac{17}{4} arctan (\frac{x}{4}) + C$ $17/4 arctan(x/4)+C$ . Standard form: $\int \frac{1}{a^{2} + x^{2}} d x = (\frac{1}{a}) arctan (\frac{x}{a}) + C$ $int1/(a^2+x^2)dx=(1/a)arctan (x/a)+C$

Explanation:

Alternatively substitute $x = 4 tan u$ $x=4tan u$ and use $1 + {tan}^{2} u = {sec}^{2} u$ $1+tan^2u=sec^2u$ giving the integral as $\int (17 \times 4) \frac{{sec}^{2} u}{16 {sec}^{2} u} d u$ $int (17 xx 4)sec^2 u/(16 sec^2 u)du$ and hence the answer.

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How do you find the integral of $\int \frac{17}{16 + x^{2}} d x$ $int 17/(16+x^2)dx$ ?

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How do you find the integral of int 17/(16+x^2)dx∫1716+x2dxint 17/(16+x^2)dx?

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How do you find the integral of $\int \frac{17}{16 + x^{2}} d x$ $int 17/(16+x^2)dx$ ?