Question

Use a computer algebra system to evaluate the integral. Compare the answer with the result using table. If the answer is not the same show that they are equivalent.

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

Short answer

Using ‘CAS,we get ,

\(\int {{x^2}\sqrt {{x^2} + 4} } dx\)=\(\frac{x}{4}\left( {{x^2} + 2} \right)\left( {\sqrt {{x^2} + \left( {{2^2}} \right)} - 2\ln \left( {x + \sqrt {{x^2} + } 4} \right. + c} \right.\) \(\frac{{\sqrt {{x^2} + 4} ({x^3} + 2x - 8\ln (1\sqrt {{x^2} + 4} + x}}{4}\)

Step-by-step solution

 Calculate of integral using any computer software

\(\int {{x^2}\sqrt {{x^2} + 4} } dx\) = \(\frac{{\sqrt {{x^2} + 4} ({x^3} + 2x - 8\ln (1\sqrt {{x^2} + 4} + x}}{4}\)

 Step 2:  Integration using tables :

We know the formula:

\(\int {{u^2}\sqrt {{a^2} + {u^2}} du = \frac{4}{8}} \left( {{a^2} + 2{u^2}} \right)\sqrt {{a^2} + {u^2}} - \frac{{a4}}{8}\ln \left( {4 + \sqrt {{a^2} + {u^2}} } \right) + c\)

Here , u=x and a=2

: \(\int {\left( {{x^2}} \right)} \sqrt {{x^2} + \left( {{2^{^2}}} \right)} = \frac{x}{8}\left( {4 + 2{x^2}} \right)\sqrt {{x^2} + 4} - \frac{{16}}{8}\ln \left( {x + \sqrt {4 + {x^2}} } \right) + c\)

= \(\frac{x}{4}\left( {{x^2} + 2} \right)\left( {\sqrt {{x^2} + \left( {{2^2}} \right)} - 2\ln \left( {x + \sqrt {{x^2} + } 4} \right. + c} \right.\)

Hence, we see both answer are equivalent.