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First, you use the substitution u=1+e^x which implies that du=e^x dx. Then, you factorise e^3x = e^2x e^x and replace e^x dx with du. Then by re...

There are two main methods to solve simultaneous equations. Elimination and Substitution. Usually one is easier and quicker than the other, but this is dependent upon the question you face in the exam so ...

First, use the substitution u=e^x (which implies dx=du/u) to make the integral ∫6/(u*(u+2)))du. Next seperate the fraction using partial fractions and expand to form 3∫1/u du - 3∫1/(u+2) du. Next integrat...

Firstly, we need to eliminate on of the terms of the equation, either the x or the y term, by combining the two equations into one equation. To do this we can add or subtract the two equations from each o...

This is a very cunning application of the integration by parts rule. Although it might look at first like integration by parts doesn't apply here since there is only the one factor, there is actually a hi...