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$\int\frac{1}{(x+5)\sqrt{x+4}}dx$ is: |
$\tan^{-1}\sqrt{x+4}+C$ $2\tan^{-1}\sqrt{x+4}+C$ $-\tan^{-1}\sqrt{x+4}+C$ $-2\tan^{-1}\sqrt{x+4}+C$ |
$2\tan^{-1}\sqrt{x+4}+C$ |
$I=\int\frac{1}{(x+5)\sqrt{x+4}}dx$. Put $x + 4 = t^2$ to get $dx = 2t\, dt$ $⇒I=\int\frac{2t\,dt}{(t^2+1)t}=2\int\frac{dt}{t^2+1}=2tan^{-1}(t)+c=2tan^{-1}(\sqrt{x+4})+c$ |
