I=∫ (dx/x^2 16) =∫ #160; (1/x^2 4^2)dx >(in>the>form>of>(1/x^2 a^2)) = (1/2× #10;4)log | (x 4/x+4) |+c =(1/8)log | (x ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Functions

∫dx/x2-16

Question

$\int \frac{d x}{x^{2} - 16}$

Open in App

Solution

$I = \int (\frac{d x}{x^{2} - 16}) = \int (\frac{1}{x^{2} - 4^{2}}) d x (i n t h e f o r m o f (\frac{1}{x^{2} - a^{2}})) = (\frac{1}{2 \times 4}) l o g ∣ ∣ (\frac{x - 4}{x + 4}) ∣ ∣ + c = (\frac{1}{8}) l o g ∣ ∣ (\frac{x - 4}{x + 4}) ∣ ∣ + c$

Suggest Corrections

Similar questions

\frac{d}{d x} [\sqrt{16 - x^{2}}]

\int \frac{d x}{x^{2} + 16}

$\int \frac{d x}{x^{2} \sqrt{16 - x^{2}}}$ has the value equal to

y = \frac{sin 4 x}{x^{2} + 16}

\frac{d y}{d x}

If \infty \int 0 \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2}) (x^{2} + c^{2})} d x = \frac{π}{2 (a + b) (b + c) (c + a)} then, \infty \int 0 \frac{d x}{(x^{2} + 1) (x^{2} + 16)} =

Join BYJU'S Learning Program