Question

Solve $\int \frac{dx}{\sqrt{x}+\sqrt[4]{4x}}$

• A. $2\sqrt{x}+4\sqrt[4]{4x}+4\log (1+\sqrt[4]{4x})+c$
• B. $2\sqrt{x}+2\sqrt[4]{4x}+4\log (1+\sqrt[4]{4x})+c$
• C. $2\sqrt{x}-4\sqrt[4]{4x}+4\log (1+\sqrt[4]{4x})+c$
• D. $\sqrt{x}+3\sqrt[4]{4x}-2\log (x+1)+c$

Answer 1

The correct option is C $2\sqrt{x}-4\sqrt[4]{4x}+4\log (1+\sqrt[4]{4x})+c$

$\int \frac{dx}{\sqrt{x}+\sqrt[4]{4x}}$
$=\int \frac{dx}{\sqrt{x}(1+\frac{1}{\sqrt[4]{4x}})}$
Put $\sqrt[4]{4x}=t$, $x=t^4$

$\Rightarrow dx=4t^{3}dt$
So, $I=4\int \frac{t^2}{t+1}dt$

$=4\int \frac{t(t+1)-t}{t+1}dt$

$=4\int tdt-4\int \frac{t}{t+1}dt$

$=4[\frac{t^2}{2}-\int (1-\frac{1}{t+1}\left)dt\right]$

$=4[\frac{t^2}{2}-t+\log (t+1)+C]$

Now, Put the value of t, We get,

$I=2\sqrt{x}-4\sqrt[4]{4x}+4\log (1+\sqrt[4]{4x})+C$