 cases, you can find the integral by using the special integral. It is not factor x minus 6 plus b, but both the sides. So 4 here and 2 over f which implies a is 2, so 3 is equal to minus 12 plus b, b is equal to 15. So 4x plus 3 can be written as 2 times 2x minus 6 plus 15. Now you replace this expression in terms, in place of the numerator which I am going to do it here. So when you replace it you get 2 times 2x minus 6 plus 15 by x square minus 6x plus 4 dx, which becomes 2 times 2x minus 6 by x square minus 6x plus 4 plus 15 times dx by x square minus 6x plus 4. Now in this if I take x square minus 6x plus 4 as t, you can easily see 2x minus 6 dx will become dt. So this integral converts to 2 dt whereas in this case I can complete the square in the t whole square minus 5. Minus 5 you can make it as, okay. So the answer finally becomes 2LnT. t in this case is the quadratic here x square minus 6x plus 4 and in this case I will have 15 times. This is following the formula of x square minus a square. So which is 1 by 2a ln x minus a by, the procedure begins with this step which is the most important. If you are able to recall that I have to use this step then after that things become pretty easy. Yes, get that idea. See Vishak Darya has to be simple as that. Find a and b like a very coefficient of x and cos square. This is all right. Let's take, try this out under root of 6 minus sin x t. Yes, correct. Then you break. Okay. Now in the denominator you convert this cos square as 1 minus sin x t. Now the moment you get simply substitute sin x as t that is cos x dx will become dt. That is denominator will become 4t minus 1 dt. Denominator will become t square minus 4t plus 5 minus 1 as 8 times the derivative. 4 implies a and minus 1 is minus 4a plus b. So when a is 2b becomes 7 correct. So I can write this as 2 times 2t minus 4 plus 7. So 2 times integral of 2t minus 4 by under root t square minus 14 plus 4 under root t square minus 14 plus 5. Let's take it as k. So 2 times dk by under root t square minus 14 plus 5. So k to the power 2 root k. So it will be 4 root t square minus 14 plus 5. Correct. Put your k. k is under root t square. So sin square x minus 4 sin x. Plus 5 plus 7 as sin square x minus 4 sin x plus 5. So these are all pretty easy to manage. You just don't make any mistakes in finding your a out and b out. And of course do not apply. Arti Shalmaras ensures that you solve at least 10, 15 of these things. Integrals of the type 1 in this case and cos square will become secant square x dx. Denominate tan square plus 9. That says t that is the second step. So your secant square x dx will become dt. That is your integral will convert to 4 common out. No. You may choose to. You may choose. Just following size method. You can write this also. 1 sixth tan inverse 2 tan x by 3. Which becomes 1 by 6 tan inverse 2 t by 3. Which gives you the same answer as what we got here. Up to you what which method you want to follow. You can multiply x square x plus cos square x to 3. 3 to 1. So it is dt by 3 minus t minus t square between a minus x. So it is x square. x plus x square. x plus x square. That is correct. x minus a by x plus a is when it is x square minus a square. But a square minus x square it is. It is a loss of 1. It is a loss of 1. A square minus x square x plus a by x. 1 by x. But it does not matter even if you stop the positions. 1 by 2 by 2 by 2 by 3. Sir. L n of 2 minus 1 by 2 by 3. That is minus 2 by 3 by tan x by 2. That is correct. That is correct.