Question

How do you find the integral of $\sqrt {{x^2} + 9} dx$?

Answer 1

The above question is based on the concept of integration. Since it is an indefinite integral which has no upper and lower limits, we can apply integration properties by integrating it in parts so that we can find the antiderivative of the above equation.

Complete step by step solution:
Integration is a way of finding the antiderivative of any function. It is the inverse of differentiation. It denotes the summation of discrete data.

Calculation of small problems is an easy task but for adding big problems which include higher limits, integration method is used. The above given equation is an indefinite integral which means there are no upper or lower limits given.

The above equation should be in the below form.
$\int {f(x) = F(x) + C}$
where C is constant.

So, we equate the integral with $t + x$

Then rearranging it and bringing $x$ on Left hand side and all other terms on right hand side.

$$\sqrt {{x^2} + 9} = t + x \\\ x = \dfrac{{9 - {t^2}}}{{2 \times t}} \\\$$

Further dx will be,
$dx = - \dfrac{{{t^2} + 9}}{{2{t^2}}}dt$

So, by substituting in the main equation we get,

$$\Rightarrow - \dfrac{1}{2}\int {\dfrac{{{{({t^2} + 9)}^2}}}{{{t^3}}}} \\\ \Rightarrow - \dfrac{1}{2}\int {\left( {t + \dfrac{{18}}{t} + \dfrac{{81}}{{{t^3}}}} \right)} dt \\\$$

Now by integrating the terms we get,
$- \dfrac{1}{2}\left( {\dfrac{{{t^2}}}{2} + 18\ln (t) - \dfrac{{81}}{{2{t^2}}}} \right) + C$

Since integration of $\dfrac{1}{x}$ is natural log(ln) and the power in the denominator if taken, the numerator becomes negative so when we integrate it by 1 it gets subtracted and reduced from3 to 2.

Note: The above method can also be solved in another alternative way by using integration by parts. The formula for integration by parts is $\int {uvdx = u\int {vdx - \int {\left( {u'\int {vdx} } \right)dx} } }$
where we can be considered as $\sqrt {{x^2} + 9}$ and v can be considered as 1. $\left( {I = \int {\sqrt {{x^2} + 9} \times 1dx} } \right)$