What is $\int \frac{23}{\sqrt{13 + 3 x^{2}}} d x$ $int 23/sqrt(13+3x^2) dx$ ?

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Answer 1 · 2015-11-12T13:01:59

$23 \int \frac{1}{\sqrt{13 + 3 x^{2}}} d x$ $23int1/sqrt(13+3x^2)dx$

$x = \sqrt{\frac{13}{3}} u$ $x = sqrt(13/3)u$

$d x = \sqrt{\frac{13}{3}} d u$ $dx = sqrt(13/3)du$

$u = \sqrt{\frac{3}{13}} x$ $u = sqrt(3/13)x$

$23 \sqrt{\frac{13}{3}} \int \frac{1}{\sqrt{13 + 13 u^{2}}} d u$ $23sqrt(13/3)int 1/sqrt(13+13u^2)du$

$23 \sqrt{\frac{13}{3}} \cdot \sqrt{\frac{1}{13}} \int \frac{1}{\sqrt{1 + u^{2}}} d u$ $23sqrt(13/3)*sqrt(1/13)int 1/sqrt(1+u^2)du$

$23 \sqrt{\frac{1}{3}} [arcsin h (u)] + C$ $23sqrt(1/3)[arcsinh(u)]+C$

Substitute back

$23 \sqrt{\frac{1}{3}} [arcsin h (\sqrt{\frac{3}{13}} x)] + C$ $23sqrt(1/3)[arcsinh(sqrt(3/13)x)]+C$

What is int 23/sqrt(13+3x^2) dx∫23√13+3x2dxint 23/sqrt(13+3x^2) dx?