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Write the value of $\int \large\frac{dx}{x^2+16}$

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• $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan^{-1}\big(\large\frac{x}{a}\big)+c I=\int\large\frac{dx}{x^2+16} This is of the form \int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan^{-1}\big(\large\frac{x}{a}\big)+c$
On integrating we get,
$I=\large\frac{1}{4}$$\tan^{-1}\big(\large\frac{x}{4}\big)$$+c$
answered Dec 2, 2013