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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{3\text{x}+5}{\text{x}^3-\text{x}^2-\text{x}+1}\ \text{dx}$
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Answer

To evaluate the integrals follow the steps:
$\int\frac{3\text{x}+5}{\text{x}^3-\text{x}^2-\text{x}+1}\ \text{dx}$
Let $\frac{3\text{x}-5}{(\text{x}-1)^2(\text{x}+1)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{(\text{x}-1)^2}+\frac{\text{C}}{\text{x}+1}$
$3\text{x}+5=\text{A}(\text{x}-1)(\text{x}+1)+\text{B}(\text{x}+1)+\text{C}(\text{x}-1)^2$
For x = 1 B = 4
$\text{For x} = -1\ \text{C}=\frac{1}{2}$
$\text{For x} = 0\ \text{A}=-\frac{1}{2}$
Therefore
$\int\frac{3\text{x}+5}{(\text{x}-1)^2(\text{x}+1)}\ \text{dx}=-\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}$
$=-\frac{1}{2}\ln|(\text{x}-1)|-\frac{4}{(\text{x}-1)}+\frac{1}{2}\ln|(\text{x}+1)|+\text{C}$
$=\frac{1}{2}\ln\big|\frac{\text{x}+1}{\text{x}-1}\big|-\frac{4}{(\text{x}-1)}+\text{C}$

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