Question

How do you find the antiderivative of $\dfrac{1}{{{x^2} + 10x + 29}}dx$ ?

Answer 1

Hint : We need to evaluate $\int {\dfrac{1}{{{x^2} + 10x + 29}}dx}$ , that is antiderivative means integration. We know that the term inside the integral sign is called integrand. In the integrand we have a quadratic equation, we first find the factors for that equation then we solve this easily by using partitions. If the factors are complex then we make use of completing the square method.

Complete step-by-step answer :
Given,
$\int {\dfrac{1}{{{x^2} + 10x + 29}}dx} {\text{ }} - - - - (1)$
Now consider
${x^2} + 10x + 29$
Now we have $\Delta = {b^2} - 4ac$ . If $\Delta < 0$ then roots are complex, if $\Delta > 0$ then the roots are real.
We have $a = 1,b = 10$ and $c = 29$ . Substituting we have
$\Delta = {(10)^2} - 4(1)(29) = 100 - 116 = - 16 < 0$
Hence the roots are complex.
Let’s complete the square,
Given, ${x^2} + 10x + 29$
We can write $29 = 25 + 4$
${x^2} + 10x + 29 = \left( {{x^2} + 10x + 25} \right) + 4$
We can see that in the brackets the expression is of the form ${(a + b)^2} = {a^2} + 2ab + {b^2}$ , where $a = x$ and $b = 5$ ,
${x^2} + 10x + 29 = {\left( {x + 5} \right)^2} + 4$
Then equation (1) becomes,
$\int {\dfrac{1}{{{x^2} + 10x + 29}}dx} = \int {\dfrac{1}{{{{\left( {x + 5} \right)}^2} + 4}}dx}$
$= \int {\dfrac{1}{{{{\left( {x + 5} \right)}^2} + 4}}dx}$
Now we make substitute $x + 5 = t$
Differentiating with respect to ’x’ we have,
$dx = dt$
Then,
$= \int {\dfrac{1}{{{t^2} + 4}}dt}$
We can rewrite it as
$= \int {\dfrac{1}{{{t^2} + {2^2}}}dt}$
We know the formula, that is $\int {\dfrac{1}{{{t^2} + {a^2}}}dt} = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{t}{a}} \right) + c$ , applying this we have,
$= \dfrac{1}{2}{\tan ^{ - 1}}\left( {\dfrac{t}{2}} \right) + c$ , where ‘c’ is the integration constant.
But we need the answer in terms of ‘x’ only.
We have $x + 5 = t$ then we have,
$= \dfrac{1}{2}{\tan ^{ - 1}}\left( {\dfrac{{x + 5}}{2}} \right) + c$
Thus we have,
$\int {\dfrac{1}{{{x^2} + 10x + 29}}dx} = = \dfrac{1}{2}{\tan ^{ - 1}}\left( {\dfrac{{x + 5}}{2}} \right) + c$ , where ‘c’ is the integrating constant.
So, the correct answer is “$\dfrac{1}{2}{\tan ^{ - 1}}\left( {\dfrac{{x + 5}}{2}} \right) + c$ ”.

Note : Here we have an indefinite integral that is no upper limit and lower limit. Hence, in the case of indefinite integral we have integration constant. In definite integral we have lower limits and upper limits. Hence, in the case of definite integral we don’t have integration constant. As we can see in the above problem by using the substitute rule we can simplify the problem easily. We can solve the above integral without substitution also.