Question

$\int \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}$ $dx$ is equal to

• A. $\frac{x^2}{2}+x+c$
• B. $\frac{x^2}{2}-x+c$
• C. $\frac{x^2}{2}+c$
• D. $x+c$

Answer 1

Answer: (A)

$\frac{x^2}{2}+x+c$

$\int \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}$ $dx$

$=\int \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+x^2+6x+8}}}$ $dx$

$=\int \sqrt{1+x\sqrt{1+(x+1)\sqrt{x^2+6x+9}}}$ $dx$

$=\int \sqrt{1+x\sqrt{1+(x+1)\sqrt{(x+3{)}^2}}}$ $dx$

$=\int \sqrt{1+x\sqrt{1+(x+1)(x+3)}}$ $dx$

$=\int \sqrt{1+x\sqrt{1+x^2+4x+3}}$ $dx$

$=\int \sqrt{1+x\sqrt{x^2+4x+4}}$ $dx$

$=\int \sqrt{1+x\sqrt{(x+2{)}^2}}$ $dx$

$=\int \sqrt{1+x(x+2)}$ $dx$

$=\int \sqrt{1+x^2+2x}$ $dx$

$=\int \sqrt{(x+1{)}^2}$ $dx$

$=\int (x+1)$ $dx$

$=\int x$ $dx$ $+\int 1$$dx$

$=\frac{x^2}{2}+x+c$

Ans is $Opt:\left[A\right]$