 Hello friends and how are you all today? The question says integrate the following rational function and the function which is given to us is 1 upon x into x raised to the power 4 minus 1. Now to make our procedure simplified, we can multiply the numerator and the denominator by 4x cube because we know that if we take the value of x raised to the power 4 equal to t then its derivative will be equal to 4x cube dx and that will make the whole procedure simplified. So we have it equal to 4x cube upon 4x raised to the power 4 that is the exponent of x is 1. So when the exponents are, when base are same then the exponents get added. So we have x raised to the power 4 into x raised to the power 4 minus 1. Now integrating both sides that is the original function that we had can be written as integral 4x cube upon 4x raised to the power 4 into x raised to the power 4 minus 1 dx. Now put x raised to the power 4 equal to t then we have 4 into x raised to the power 3 dx is equal to dt. So we can write dt upon, let's take the constant out of the integration. So we have 1 upon 4 integral dt upon t into t minus 1. Now decomposing this rational fraction into partial fractions we have 1 upon t into t minus 1 can be written as 1 upon t sorry a upon t plus b upon t minus 1. Now taking the LCM and equating the numerators we will have 1 is equal to a into t minus 1 plus b into t. Now equating the coefficients of the variable t we have a plus b is equal to 0 and equating the constants we have minus a is equal to 1. This implies the value of a will be equal to minus 1 and that means on substituting the value of a as minus 1 in this equation we have the value of b equal to 1. So now we can write this fraction as or this integral as 1 upon 4 integral minus 1 upon t plus 1 upon t minus 1 into dt. So further on separating the integration sign we have 1 upon 4 minus 1 upon 4 integral dt upon t plus 1 upon 4 integral dt upon t minus 1. Now the answer will be 1 upon 4 log mod t plus 1 upon 4 log mod t minus 1 plus c. Now on substituting the value of t as x raise to the power 4 we have 1 upon minus 1 upon 4 log. Now this we can write as say minus 1 upon t so that minus sign will get cancelled plus c. So we have 1 upon 4 log x raise to the power 4 minus 1 upon x raise to the power 4 plus c and this is the required answer to the session. This completes the question that was given to us hope you enjoyed and understood the whole concept well. Remember you need to decompose the rational fraction into partial fractions and use your previous knowledge to simplify it further. Have a nice day.