Question

Find the value of the given integral.
$\int {\dfrac{{{{\text{x}}^2}{\text{ + 4x}}}}{{{{\text{x}}^3}{\text{ + 6}}{{\text{x}}^2}{\text{ + 5}}}}{\text{dx}}}$

Answer 1

Hint: Let us substitute in the given integral to find the value of the given integral easily. So, let us assume denominator as t and then substitute numerator in terms of dt.

Complete step-by-step answer:
Now, we will use the substitution technique. We will let the denominator term as t and then differentiate the denominator term because the given integral is in proper form i.e. the degree of numerator is less than degree of denominator. So,
Let ${{\text{x}}^3}{\text{ + 6}}{{\text{x}}^2}{\text{ + 5 = t}}$
Differentiating both sides with respect to x, we get
$(3{{\text{x}}^2}{\text{ + 12x)dx = dt }}$
$({{\text{x}}^2}{\text{ + }}{\text{ 4x)dx = }}\dfrac{{{\text{dt}}}}{3}$
Substituting the value of t and dx in the given integral, we get
$\int {\dfrac{{{{\text{x}}^2}{\text{ + 4x}}}}{{{{\text{x}}^3}{\text{ + 6}}{{\text{x}}^2}{\text{ + 5}}}}{\text{dx}}}$ =
Now, $\int {\dfrac{{{\text{dx}}}}{{\text{x}}}{\text{ = ln}}\left| {\text{x}} \right|}$
So, $\int {\dfrac{{{{\text{x}}^2}{\text{ + 4x}}}}{{{{\text{x}}^3}{\text{ + 6}}{{\text{x}}^2}{\text{ + 5}}}}{\text{dx}}}$ = $\dfrac{{{\text{ln}}\left| {\text{t}} \right|}}{3}{\text{ + c}}$ , where c is the integration constant.
Now putting the value of t in the above equation, we get
$\int {\dfrac{{{{\text{x}}^2}{\text{ + 4x}}}}{{{{\text{x}}^3}{\text{ + 6}}{{\text{x}}^2}{\text{ + 5}}}}{\text{dx}}}$ = $\dfrac{{{\text{ln}}\left| {{{\text{x}}^3}{\text{ + }}{\text{ 6}}{{\text{x}}^2}{\text{ + 5}}} \right|}}{3}{\text{ + c}}$
So, the given integral has the value $\dfrac{{{\text{ln}}\left| {{{\text{x}}^3}{\text{ + }}{\text{ 6}}{{\text{x}}^2}{\text{ + 5}}} \right|}}{3}{\text{ + c}}$.

Note: While solving questions which include integration of given terms, we have to check whether the given integral is proper or improper. In a proper integral, the degree of the numerator is less than that of the denominator and vice – versa in the improper integral. Also, we have to write the integration constant c when we are dealing with indefinite integrals i.e. integrals with no limit.