Question

make substitution to express the integrand a rational function and them evaluate the integral.

\(\int\limits_{1/3}^3 {\frac{{\sqrt x }}{{{x^2} + x}}} dx\)

Short answer

The value of the given integral is shown below.

\(\int\limits_{1/3}^3 {\frac{{\sqrt x }}{{{x^2} + x}}} dx\)=\(2{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) + c\)

Step-by-step solution

Rewrite the integral & evaluate it.

let u=\(\sqrt 3 \)=u²=x : du(2u) =dx & limit will be

\(\int\limits_{1/3}^3 {\frac{{\sqrt x }}{{{x^2} + x}}} dx\)=2\(\int\limits_{1/\sqrt 3 }^{\sqrt 3 } {\frac{{{u^2}}}{{{u^4} + {u^2}}}} du\) \(x = 1/3to3sou \to {1 \mathord{\left/

{\vphantom {1 {\sqrt 3 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 3 }}to\sqrt 3 \)

=\(2\int\limits_{1/3}^{\sqrt 3 } {\frac{1}{{{u^2} + 1}}} du\)

:- \(\int\limits_{1/3}^3 {\frac{{\sqrt x }}{{{x^2} + x}}} dx\)=2\(\left( {{{\tan }^{ - 1}}\left( u \right)} \right)_{1/\sqrt 3 }^{\sqrt 3 }\)\(\)

Apply the limit and substitute 

u=\(\sqrt x \) \(\int\limits_{1/3}^3 {\frac{{\sqrt x }}{{{x^2} + x}}} dx\)\( = 2\left( {{{\tan }^{ - 1}}\left( {\sqrt 3 } \right) - {{\tan }^{ - 1}}\left( {1/\sqrt 3 } \right)} \right)\)

\( = 2{\tan ^{ - 1}}\sqrt 3 - 2{\tan ^{ - 1}}\left( {1/\sqrt 3 } \right) + c\)

Hence, the solution of the given integral can be given as

\(\int\limits_{1/3}^3 {\frac{{\sqrt x }}{{{x^2} + x}}} dx\)\( = 2{\tan ^{ - 1}}\sqrt 3 - 2{\tan ^{ - 1}}1/\sqrt 3 + c\)