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# Simple Integration 1.02

For any natural number m,evaluate

$$\int (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^m+6)^{1/m}$$dx

Note by Anirudha Nayak
3 years ago

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Rewrite as : $\frac{1}{6 m} \int (6mx^{3m-1}+6mx^{2m-1}+6mx^{m-1} )(2x^{3m}+3x^{2m}+6x^m)^{1/m} \ \mathrm{d}x.$ Note that this the integrand has the form of : $$u'(x) [u(x)]^{1/m}$$, where $$u(x)=(2x^{3m}+3x^{2m}+6x^m)^{1/m}.$$

Then the integral is : $\frac{1}{6m} \cdot \left(\frac{1}{m}+1\right)^{-1} \cdot [u(x)]^{1/m+1} = \frac{\left(2x^{3m}+3x^{2m}+6x^m\right)^{m+1/m}}{6(m+1)}.$ · 3 years ago

This question is from IIT JEE. First multiply and divide by x i.e. divide the first term by x and multiply the 2nd term by x (which becomes x^m on going inside the polynomial). Now you can easily solve the integral by taking the new 2nd term as a variable say 't'. · 3 years ago

i gave this one, as usually as one sees the question he would take x^m common from the first part and than multiply inside the bracket and hence he will be TRAPPED and couldnt solve or if try it will be long compared to the way mentioned...................... · 3 years ago