Solveeit Logo
Login

Question

Question: \[\int_{}^{}\sqrt{x^{2} + 8x + 12}dx\]...

∫x2+8x+12dx\int_{}^{}\sqrt{x^{2} + 8x + 12}dx

A

12(x+4)x2+8x+12+2log⁡∣x+4+x2+8x+12∣+c\frac{1}{2}(x + 4)\sqrt{x^{2} + 8x + 12} + 2\log|x + 4 + \sqrt{x^{2} + 8x + 12}| + c

B

12(x+4)x2+8x+12−2log⁡∣x+4+x2+8x+12∣+c\frac{1}{2}(x + 4)\sqrt{x^{2} + 8x + 12} - 2\log|x + 4 + \sqrt{x^{2} + 8x + 12}| + c

C

(x+4)x2+8x+12+log⁡∣x+4+x2+8x+12∣+c(x + 4)\sqrt{x^{2} + 8x + 12} + \log|x + 4 + \sqrt{x^{2} + 8x + 12}| + c

D

(x+4)x2+8x+12−log⁡∣x+4+x2+8x+12∣+c(x + 4)\sqrt{x^{2} + 8x + 12} - \log|x + 4 + \sqrt{x^{2} + 8x + 12}| + c

Answer

12(x+4)x2+8x+12−2log⁡∣x+4+x2+8x+12∣+c\frac{1}{2}(x + 4)\sqrt{x^{2} + 8x + 12} - 2\log|x + 4 + \sqrt{x^{2} + 8x + 12}| + c

Explanation

Solution

Let I=∫x2+8x+12dx=∫(x2+8x+16)−4dx\mathbf{I =}\int_{}^{}\sqrt{\mathbf{x}^{\mathbf{2}}\mathbf{+ 8x + 12}}\mathbf{dx =}\int_{}^{}\sqrt{\mathbf{(}\mathbf{x}^{\mathbf{2}}\mathbf{+ 8x + 16)}\mathbf{-}\mathbf{4}}\mathbf{dx}

=∫(x+4)2−22dx=∫t2−22dx=∫t2−22dt.= \int_{}^{}\sqrt{(x + 4)^{2} - 2^{2}}dx = \int_{}^{}\sqrt{t^{2} - 2^{2}}dx = \int_{}^{}\sqrt{t^{2} - 2^{2}}dt. (putting x+4=t⇒dx=dt)x + 4 = t \Rightarrow dx = dt)

=12tt2−22−12.22log⁡∣t+t2−22∣+c= \frac{1}{2}t\sqrt{t^{2} - 2^{2}} - \frac{1}{2}.2^{2}\log|t + \sqrt{t^{2} - 2^{2}}| + c

=12(x+4)(x+4)2−4−2log⁡∣x+4+(x+4)2−4∣+c= \frac{1}{2}(x + 4)\sqrt{(x + 4)^{2} - 4} - 2\log|x + 4 + \sqrt{(x + 4)^{2} - 4}| + c

=12(x+4)x2+8x+12−2log⁡∣x+4+x2+8x+12∣+c= \frac{1}{2}(x + 4)\sqrt{x^{2} + 8x + 12} - 2\log|x + 4 + \sqrt{x^{2} + 8x + 12}| + c