What is the integration of \[ \int {\frac{{{x^{1/2}}}}{{1 + {x^{3/4}}}}} dx\] ~~~~~[NCERT Exemp. Q. 12,Page 164 ]

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

Let’s put \[ x = {t^4} \Rightarrow dx = 4{t^3}dt\]
therefore,\[I = 4\int {\frac{{{t^2}\left( {{t^3}} \right)}}{{1 + {t^3}}}} dt = 4\int {\left( {{t^2} – \frac{{{t^2}}}{{1 + {t^3}}}} \right)} dt\]

\[I = 4\int {{t^2}} dt – 4\int {\frac{{{t^2}}}{{1 + {t^3}}}} dt\]

\[I = {I_1} – {I_2}\]

\[{I_1} = 4\int {{t^2}} dt = 4 \cdot \frac{{{t^3}}}{3} + {C_1} = \frac{4}{3}{x^{3/4}} + {C_1}\]

Now,\[{I_2} = 4\int {\frac{{{t^2}}}{{1 + {t^3}}}} dt\]


Again, Let’s put \[ 1 + {t^3} = z \Rightarrow 3{t^2}dt = dz\]


\[ \Rightarrow \] \[{t^2}dt = \frac{1}{3}dz = \frac{4}{3}\int {\frac{1}{z}} dz\]

\[ = \frac{4}{3}\log |z| + {C_2} = \frac{4}{3}\log \left| {\left( {1 + {t^3}} \right)} \right| + {C_2}\]

\[ = \frac{4}{3}\log \left| {\left( {1 + {x^{3/4}}} \right)} \right| + {C_2}\]

therefore,\[I = \frac{4}{3}{x^{3/4}} + {C_1} – \frac{4}{3}\log \left| {\left( {1 + {x^{3/4}}} \right)} \right| – {C_2}\]


\[ = \frac{4}{3}{x^{3/4}} – \log \mid \left( {1 + {x^{3/4}}} \right) + C\]

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/