Question

Evaluate the integral if exists.

\(\int {\frac{{{\rm{x + 2}}}}{{\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + 4x}}} }}} {\rm{dx}}\)

Short answer

The required solution is\(\int {\frac{{x + 2}}{{\sqrt {{x^2} + 4x} }}} dx = \sqrt {{x^2} + 4x} + C\).

Step-by-step solution

Methods of evaluating integral

For evaluating integrals there are various methods like substitution (also called u-substitution), integration by parts, partial fractions etc.

Applying substitution method

Let \(u = {x^2} + 4x\)

\(\begin{aligned}{c}du &= (2x + 4)dx\\\frac{{du}}{2} &= (x + 2)dx\end{aligned}\)

Substituting u in given function to get required results

Substitute \(u = {x^2}{\rm{ + 4x and }}\frac{{du}}{2} = (x + 2)dx\)

\(\begin{aligned}{c}\int {\frac{{x + 2}}{{\sqrt {{x^2} + 4x} }}} dx &= \int {\frac{1}{{\sqrt u }}} \frac{{du}}{2}\\ &= \frac{1}{2} \cdot \int {{u^{ - 1/2}}} du\\ &= \frac{1}{2} \cdot \frac{{{u^{1/2}}}}{{1/2}}\\ &= {u^{1/2}} + C\\ &= \sqrt u + C\end{aligned}\)

Replace value of u with original value

\(\int {\frac{{x + 2}}{{\sqrt {{x^2} + 4x} }}} dx = \sqrt {{x^2} + 4x} + C\)

Therefore, \(\int {\frac{{x + 2}}{{\sqrt {{x^2} + 4x} }}} dx = \sqrt {{x^2} + 4x} + C\).