How do you integrate $\dfrac{{3x + 5}}{{{x^2} + 4x + 13}}$ using partial fractions?
Answer
603.3k+ views
Hint: In this question we are asked to integrate using partial fraction. In order to proceed with this question we need to know what integration and partial fraction is. The integration denotes the summation of the data which is discrete. The integral is calculated to find the functions, which cannot be measured singularly, which will describe the area, displacement, volume, that occurs due to a collection of small data. In a broad sense, in calculus, algebra and geometry are implemented, the idea of limit is used where. Limit tells us the result of points on a graph in the way as how they get closer to each other until their distance comes down to zero. An algebraic fraction can be broken down into simpler parts which are known as “partial fractions“.
Complete step by step solution:
We are given,
$\dfrac{{3x + 5}}{{{x^2} + 4x + 13}}$
The fraction cannot be integrated using partial fraction because the quadratic equation in the denominator cannot be factored i.e. it does not have any real roots because discriminant<0.
$
\Rightarrow {b^2} - 4ac \\
\Rightarrow {4^2} - 4 \times 1 \times 13 \\
\Rightarrow 16 - 52 \\
\Rightarrow - 36 \;
$
Hence, it cannot be solved by partial fraction.
Note: Some rules to use partial fraction-
I.The denominator must be of at least one degree more than the numerator.
II.If a linear factor in the denominator is repeated n times, there will be n corresponding partial fractions with degree 1 to n.
III.The denominator of the fraction should have real roots. It should be able to be factored and the discriminant should be equal to or more than zero
Complete step by step solution:
We are given,
$\dfrac{{3x + 5}}{{{x^2} + 4x + 13}}$
The fraction cannot be integrated using partial fraction because the quadratic equation in the denominator cannot be factored i.e. it does not have any real roots because discriminant<0.
$
\Rightarrow {b^2} - 4ac \\
\Rightarrow {4^2} - 4 \times 1 \times 13 \\
\Rightarrow 16 - 52 \\
\Rightarrow - 36 \;
$
Hence, it cannot be solved by partial fraction.
Note: Some rules to use partial fraction-
I.The denominator must be of at least one degree more than the numerator.
II.If a linear factor in the denominator is repeated n times, there will be n corresponding partial fractions with degree 1 to n.
III.The denominator of the fraction should have real roots. It should be able to be factored and the discriminant should be equal to or more than zero
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