Question

Evaluate the integral\(\int {\frac{{\rm{1}}}{{\sqrt {{\rm{x + }}{{\rm{x}}^{{\rm{3/2}}}}} }}} {\rm{dx}}\).

Short answer

The integral value of the given equation is\(\int {\frac{{\rm{1}}}{{\sqrt {{\rm{x + }}{{\rm{x}}^{{\rm{3/2}}}}} }}} {\rm{dx = 4}}\sqrt {{\rm{1 + }}\sqrt {\rm{x}} } {\rm{ + c}}\).

Step-by-step solution

Expand the equation.

\(\begin{aligned}{l}{\rm{I = }}\int {\frac{{\rm{1}}}{{\sqrt {{\rm{x + }}{{\rm{x}}^{{\rm{3/2}}}}} }}} {\rm{dx}}\\{\rm{ = }}\int {\frac{{\rm{1}}}{{\sqrt {{\rm{x}}\left( {{\rm{1 + }}{{\rm{x}}^{{\rm{1/2}}}}} \right)} }}} {\rm{dx}}\\{\rm{ = }}\int {\frac{{\rm{1}}}{{\sqrt {\rm{x}} \sqrt {{\rm{1 + }}\sqrt {\rm{x}} } }}} {\rm{dx}}\end{aligned}\)

Evaluate the equation.

Let\({\rm{u = 1 + }}\sqrt {\rm{x}} \to {\rm{du = }}\frac{{{\rm{dx}}}}{{{\rm{2}}\sqrt {\rm{x}} }} \to {\rm{dx = 2}}\sqrt {\rm{x}} {\rm{ du}}\)

\(\begin{aligned}{l}{\rm{ = }}\int {\frac{{\rm{1}}}{{\sqrt {\rm{u}} }}} {\rm{ \times 2du}}\\{\rm{ = 2}}\int {{{\rm{u}}^{{\rm{ - 1/2}}}}} {\rm{du}}\\{\rm{ = 4}}\sqrt {\rm{u}} {\rm{ + c}}\end{aligned}\)

Substitute \({\rm{u = 1 + }}\sqrt {\rm{x}} \)

\({\rm{ = 4}}\sqrt {{\rm{1 + }}\sqrt {\rm{x}} } {\rm{ + c}}\)

Therefore, the integral value of the given equation is\(\int {\frac{{\rm{1}}}{{\sqrt {{\rm{x + }}{{\rm{x}}^{{\rm{3/2}}}}} }}} {\rm{dx = 4}}\sqrt {{\rm{1 + }}\sqrt {\rm{x}} } {\rm{ + c}}\).