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The value of $∫\frac{1}{\sqrt{x^2+8x+9}}dx=$ is (Given C constant of integration) |
$log[(x+4)+\sqrt{x^2+8x+9}]+C;$ $log[(x+4)-\sqrt{x^2+8x+9}]+C;$ $log[(x-4)+\sqrt{x^2-8x+9}]+C;$ $log[(x-4)-\sqrt{x^2+8x-9}]+C;$ |
$log[(x+4)+\sqrt{x^2+8x+9}]+C;$ |
The correct answer is Option (1) → $\log[(x+4)+\sqrt{x^2+8x+9}]+C$ $∫\frac{1}{\sqrt{x^2+8x+9}}dx$ $=∫\frac{1}{(x+4)^2-(\sqrt{7})^2}dx$ $=\log|(x+4)+\sqrt{x^2+8x+9}|+C$ |
