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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Integrate the function in Exercise:
$\frac{1}{\text{x}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}}$
$\Bigg[\text{Hint:}\frac{1}{\text{x}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}}=\frac{1}{\text{x}^{\frac{1}{3}}\bigg(1+\text{x}^{\frac{1}{6}}\bigg)},\text{put}\ \text{x}=\text{t}^{6}\Bigg]$
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Answer

$\frac{1}{\text{x}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}}=\frac{1}{\text{x}^{\frac{1}{3}}\bigg(1+\text{x}^{\frac{1}{6}}\bigg)}$

$\text{Let}\ \text{x}=\text{t}^{6}\Rightarrow\text{dx}=6\text{t}^{5}\text{dt}$

$\therefore\int\frac{1}{\text{x}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}}\text{dx}=\int\frac{1}{\text{x}^{\frac{1}{3}}\bigg(1+\text{x}\frac{1}{6}\bigg)}\text{dx}$

$=\int\frac{6\text{t}^{5}}{\text{t}^{2}(1+\text{t)}}\text{dt}$

$=6\int\frac{\text{t}^{3}}{(1+\text{t)}}\text{dt}$

On dividing, we obtain

$\int\frac{1}{\text{x}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}}\text{dx}=6\int\left\{(\text{t}^{2}-\text{t}+1)-\frac{1}{1+\text{t}}\right\}\text{dt}$

$=6\bigg[\bigg(\frac{\text{t}^{3}}{3}\bigg)-\bigg(\frac{\text{t}^{2}}{2}\bigg)+\text{t}-\log|1+\text{t}|\bigg]$

$=2\text{x}^{\frac{1}{2}}-3\text{x}^{\frac{1}{3}}+6\text{x}^{\frac{1}{6}}-6\log\bigg(1+\text{x}^{\frac{1}{6}}\bigg)+\text{C}$

$=2\sqrt{\text{x}}-3\text{x}^{\frac{1}{3}}+6\text{x}^{\frac{1}{6}}-6\log\bigg(1+\text{x}^{\frac{1}{6}}\bigg)+\text{C}$

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