Question

\(\int \sqrt{4^x(4^x+4)}\,dx = \cdots\)

• A. \(\frac{1}{log2}[\frac{2^x}{2}\sqrt{4^x+4}+2log|2^x+\sqrt{4^x+4}|]+c\)
• B. \(\frac{2^x}{2}\sqrt{4^x+4}+2log|2^x+\sqrt{4^x+4}|+c\)
• C. \(\frac{1}{log2}[\frac{2^x}{2}\sqrt{4^x+4}+log|2^x+\sqrt{4^x+4}|]+c\)
• D. \([\frac{2^x}{log4}\sqrt{4^x+4}+log|2^x+\sqrt{4^x+4}|]+c\)

Hint:

Write sqrt(4^x) as 2^x, substitute u = 2^x, and use the standard integral of sqrt(u^2 + a^2).

Answer 1

The Correct Option is A

Step 1: Simplify:
\(\sqrt{4^x(4^x+4)}=2^x\sqrt{4^x+4}\).

Step 2: Substitute:
Let \(u=2^x\). Then \(du=2^x\ln2\,dx\), so \(2^x\,dx=\dfrac{du}{\ln2}\). Also \(4^x=u^2\).
\[ I=\dfrac{1}{\ln2}\int\sqrt{u^2+4}\,du \]

Step 3: Use the standard result:
\(\int\sqrt{u^2+a^2}\,du=\dfrac u2\sqrt{u^2+a^2}+\dfrac{a^2}{2}\ln|u+\sqrt{u^2+a^2}|\). With \(a=2\): \(\dfrac u2\sqrt{u^2+4}+2\ln|u+\sqrt{u^2+4}|\).

Step 4: Write the final form:
\[ I=\dfrac{1}{\log2}\left[\dfrac{2^x}{2}\sqrt{4^x+4}+2\log|2^x+\sqrt{4^x+4}|\right]+c \]
Option A.

Step 5: Why the other options are wrong.
Option B omits the \(1/\ln2\) factor. Options C and D carry \(\ln|\cdot|\) with coefficient 1 instead of \(a^2/2=2\), and D also has a wrong coefficient for the first term.

Final Answer:
The integral is given by option A. \[ \boxed{\text{(A) }\dfrac{1}{\log2}\left[\dfrac{2^x}{2}\sqrt{4^x+4}+2\log|2^x+\sqrt{4^x+4}|\right]+c} \]