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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following intregals:
$\int\frac{1}{\text{x}(\text{x}^3+8)}\ \text{dx}$
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Answer

Consider the integral
$\text{I}=\int\frac{1}{\text{x}(\text{x}^3+8)}\ \text{dx}$
Rewriting the above integrade, we have,
$\text{I}=\int\frac{\text{x}^2}{\text{x}^3(\text{x}^3+8)}\ \text{dx}$
$=\frac{1}{3}\int\frac{3\text{x}^2}{\text{x}^3(\text{x}^3+8)}\ \text{dx}$
Now substracting $x^3 = t,$ we have
$3\text{x}^2\text{dx}=\text{dt}$
$\Rightarrow\text{I}=\frac{1}{3}\int\frac{\text{dt}}{\text{t}(\text{t}+8)}$
Let us seprate the integrade by partical fractions.
Thus,
$\frac{1}{\text{t}(\text{t}+8)}=\frac{\text{A}}{\text{t}}+\frac{\text{B}}{\text{t}+8}$
$\Rightarrow\frac{1}{\text{t}(\text{t}+8)}=\frac{\text{A}(\text{t}+8)+\text{Bt}}{\text{t}(\text{t}+8)}$
$\Rightarrow1=\text{A}(\text{t}+8)+\text{Bt}$
$\Rightarrow1=\text{At}+8\text{A}+\text{Bt}$
Compairing the coefficient, we have,
$\text{A}+\text{B}=0$ and $8\text{A}=1$
$\Rightarrow\text{A}=\frac{1}{8}$ and $\text{B}=-\frac{1}{8}$
Therefore,
$\text{I}=\frac{1}{3}\int\frac{\text{dt}}{\text{t}(\text{t}+8)}$
$=\frac{1}{3}\int\bigg[\frac{\frac{1}{8}}{\text{t}}-\frac{\frac{1}{8}}{\text{t}+8}\bigg]\text{dt}$
$=\frac{1}{3}\times\frac{1}{8}\int\frac{\text{dt}}{\text{t}}\text{dt}-\frac{1}{3}\times\frac{1}{8}\int\frac{\text{dt}}{\text{t}+8}$
$=\frac{1}{24}\log\text{t}-\frac{1}{24}\times\log(\text{t}+8)+\text{C}$
$=\frac{1}{24}\log\text{x}^3-\frac{1}{24}\times\log(\text{x}^3+8)+\text{C}$
$=\frac{3}{24}\log\text{x}-\frac{1}{24}\times\log(\text{x}^3+8)+\text{C}$
$=\frac{1}{8}\log\text{x}-\frac{1}{24}\times\log(\text{x}^3+8)+\text{C}$

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