Question

How do you integrate \[\int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}} \] using partial fractions?

Answer 1

Hint: We check if the degree of the numerator is strictly less than the degree of the denominator to be able to use partial fractions here. Then we use the general formula for converting the given fraction into a partial fraction. Substitute the partial fraction under the integral and solve the separate integrals.
Partial fractions is a method of dividing a single fraction into a sum of two or more fractions such that the integration becomes easier. The degree of the numerator should always be less than the degree of the denominator here. When given a fraction of the kind where the denominator is like \[{(a{x^2} + bx + c)^k}\]then we write the fraction equal to the sum \[\dfrac{{{A_1}x + {B_1}}}{{a{x^2} + bx + c}} + \dfrac{{{A_2}x + {B_2}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}} + ..... + \dfrac{{{A_k}x + {B_k}}}{{{{\left( {a{x^2} + bx + c} \right)}^k}}}\]
\[\int {\dfrac{1}{{1 + {x^2}}}dx} = {\tan ^{ - 1}}x\]

Complete step-by-step answer:
We have to integrate \[\int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}} \] using partial fractions.
Since the fraction has the degree of the denominator (4) less than the degree of the numerator (2), we can apply a method of partial fractions.
Also, the denominator of the fraction \[\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}\]is of the type \[{(a{x^2} + bx + c)^k}\], so we use the general formula given in hint to convert in to partial fraction.
We write
\[ \Rightarrow \dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}} = \dfrac{{Ax + B}}{{({x^2} + 1)}} + \dfrac{{Cx + D}}{{{{({x^2} + 1)}^2}}}\] … (1)
Take LCM in right hand side of the equation
\[ \Rightarrow \dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}} = \dfrac{{(Ax + B)({x^2} + 1) + Cx + D}}{{{{({x^2} + 1)}^2}}}\]
Cancel same denominator from both sides of the equation
\[ \Rightarrow 3{x^2} - 2x + 5 = (Ax + B)({x^2} + 1) + Cx + D\]
Open the brackets on right hand side of the equation and multiply the terms
\[ \Rightarrow 3{x^2} - 2x + 5 = A{x^3} + B{x^2} + Ax + B + Cx + D\]
Group together the coefficients with same variables on right hand side of the equation
\[ \Rightarrow 3{x^2} - 2x + 5 = A{x^3} + B{x^2} + (A + C)x + (B + D)\] … (2)
We know that two polynomials are equal if the coefficients are equal for the same variables.
Equate the coefficients on both sides of the equation (2)
\[ \Rightarrow A = 0;B = 3;A + C = - 2;B + D = 5\]
Substitute the value of A and B in to calculate values of C and D
\[ \Rightarrow A = 0;B = 3;C = - 2;D = 2\]
Substitute the values of A, B, C and D in equation (1)
\[ \Rightarrow \dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}} = \dfrac{{0.x + 3}}{{({x^2} + 1)}} + \dfrac{{ - 2x + 2}}{{{{({x^2} + 1)}^2}}}\]
\[ \Rightarrow \dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}} = \dfrac{3}{{({x^2} + 1)}} + \dfrac{{ - 2x + 2}}{{{{({x^2} + 1)}^2}}}\]
Integrate both sides of the equation with respect to x
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = \int {\dfrac{3}{{({x^2} + 1)}}dx} + \int {\dfrac{{ - 2x + 2}}{{{{({x^2} + 1)}^2}}}dx} \]
We know that we can bring out constant from the integration
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3\int {\dfrac{1}{{({x^2} + 1)}}dx} + \int {\dfrac{{ - 2x + 2}}{{{{({x^2} + 1)}^2}}}dx} \]
We know that \[\int {\dfrac{1}{{1 + {x^2}}}dx} = {\tan ^{ - 1}}x + C\] because \[\dfrac{d}{{dx}}{\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}\]
Substitute the value in right hand side of the equation
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + C + \int {\dfrac{{ - 2x + 2}}{{{{({x^2} + 1)}^2}}}dx} \]
We break the remaining integral into 2 parts
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + C - \int {\dfrac{{2x}}{{{{({x^2} + 1)}^2}}}dx} + \int {\dfrac{2}{{{{({x^2} + 1)}^2}}}dx} \]
For second integral let us substitute \[{x^2} + 1 = t\]
Differentiating both sides we get \[2xdx = dt\]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + C - \int {\dfrac{1}{{{t^2}}}dt} + \int {\dfrac{2}{{{{({x^2} + 1)}^2}}}dx} \]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + C - \int {{t^{ - 2}}dt} + \int {\dfrac{2}{{{{({x^2} + 1)}^2}}}dx} \]
Use general method of integration i.e. \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C\]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x - \left( {\dfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}}} \right) + C + \int {\dfrac{2}{{{{({x^2} + 1)}^2}}}dx} \]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x - \left( {\dfrac{{{t^{ - 1}}}}{{ - 1}}} \right) + C + \int {\dfrac{2}{{{{({x^2} + 1)}^2}}}dx} \]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{t} + C + \int {\dfrac{2}{{{{({x^2} + 1)}^2}}}dx} \]
Substitute value of \[{x^2} + 1 = t\]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{{{x^2} + 1}} + C + \int {\dfrac{2}{{{{({x^2} + 1)}^2}}}dx} \]
Now for the third integral, we substitute \[x = \tan y\]i.e. \[y = {\tan ^{ - 1}}x\]
Differentiating both sides we get \[dx = {\sec ^2}ydy\]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{{{x^2} + 1}} + C + 2\int {\dfrac{{{{\sec }^2}y}}{{{{({{\tan }^2}y + 1)}^2}}}dy} \]
We know that \[1 + {\tan ^2}y = {\sec ^2}y\]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{{{x^2} + 1}} + C + 2\int {\dfrac{{{{\sec }^2}y}}{{{{({{\sec }^2}y)}^2}}}dy} \]
Cancel same terms from numerator and denominator
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{{{x^2} + 1}} + C + 2\int {\dfrac{1}{{{{\sec }^2}y}}dy} \]
We know that \[\dfrac{1}{{\sec y}} = \cos y\]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{{{x^2} + 1}} + C + 2\int {{{\cos }^2}ydy} \]
We know \[{\cos ^2}y = \dfrac{1}{2}\left( {\cos 2y + 1} \right)\]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{{{x^2} + 1}} + C + 2 \times \dfrac{1}{2}\int {\left( {\cos 2y + 1} \right)dy} \]
Substitute \[\int {\cos ydy = - \sin y} \]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{{{x^2} + 1}} + C - \dfrac{{\sin 2y}}{2} + y\]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{{{x^2} + 1}} + C - 2\sin y\cos y + y\]
When \[x = \tan y\], then the right angle triangle with angle y will give us value of \[\sin y = \dfrac{x}{{\sqrt {1 + {x^2}} }};\cos y = \dfrac{1}{{\sqrt {1 + {x^2}} }}\]. Also put \[y = {\tan ^{ - 1}}x\]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{{{x^2} + 1}} - 2 \times \left( {\dfrac{x}{{\sqrt {1 + {x^2}} }} \times \dfrac{1}{{\sqrt {1 + {x^2}} }}} \right) + {\tan ^{ - 1}}x + C\]
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 3{\tan ^{ - 1}}x + \dfrac{1}{{{x^2} + 1}} - \dfrac{{2x}}{{1 + {x^2}}} + {\tan ^{ - 1}}x + C\]
Add like values on right hand side
\[ \Rightarrow \int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}dx} = 4{\tan ^{ - 1}}x + \dfrac{{1 - 2x}}{{{x^2} + 1}} + C\]

\[\therefore \]The value of the \[\int {\dfrac{{3{x^2} - 2x + 5}}{{{{({x^2} + 1)}^2}}}} \] using partial fractions is \[4{\tan ^{ - 1}}x + \dfrac{{1 - 2x}}{{{x^2} + 1}} + C\]

Note:
Many students make the mistake of solving the last integral wrong as they use the same formula of inverse of tangent to write the value of integral for the last integral. Keep in mind the square of the denominator prevents us from using that direct formula. Also, solve integrals separated by partial fractions one by one to avoid confusion.