What is the integration of \[\int {\frac{{\sqrt {1 + {x^2}} }}{{{x^4}}}} dx\] ~~~~~[NCERT Exemp. Q. 13,Page 164 ]

Let \[I = \int {\frac{{\sqrt {1 + {x^2}} }}{{{x^4}}}} dx = \int {\frac{{\sqrt {1 + {x^2}} }}{x}} \cdot \frac{1}{{{x^3}}}dx\]


\[ = \int {\sqrt {\frac{{1 + {x^2}}}{{{x^2}}}} } \cdot \frac{1}{{{x^3}}}dx = \int {\sqrt {\frac{1}{{{x^2}}} + 1} } \cdot \frac{1}{{{x^3}}}dx\]

Let’s put \[1 + \frac{1}{{{x^2}}} = {t^2} \Rightarrow \frac{{ – 2}}{{{x^3}}}dx = 2tdt\]

\[ \Rightarrow \] \[ – \frac{1}{{{x^3}}} = tdt\]

therefore,\[ I = – \int {{t^2}} dt = – \frac{{{t^3}}}{3} + C = – \frac{1}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{3/2}} + C\]

 

Buy Best Mathematics E-Books Visit : https://mathstudy.in/

Buy Mathematics Formula Book for Class XI,XII,JEE and other Engineering Competition Exam https://mathstudy.in/product/mathematics-formula-book/

Buy Mathematics Workbook for Class XII ( Fully Solved ) : https://mathstudy.in/product/work-book-class-xii-c-b-s-e-fully-solved/

Buy Mathematics Chapter Tests for Class XII ( Fully Solved) : https://mathstudy.in/product/mathematics-chapter-tests-class-xii-c-b-s-e/

Buy Objective Type Question Bank Class XII (Fully Solved ) : https://mathstudy.in/product/objective-type-question-bank-for-mathematics-class-xii-c-b-s-e/