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This requires your knowledge of integration by parts. The trick here is that lnx can also be written as lnx*1 (as any term multiplied by 1 is itself). We set u=lnx and dv/dx equal to 1. Hence from this we...

d[sec(x)]/dx -->d[1/cos(x)] -->( [d[1].cos(x) - d[cos(x)].1] ) / [cos(x)]^2 -->[0.cos(x) - -sin(x).1]/ [cos(x)]^2 -->sin(x)/[cos(x)]^2 -->tan(x)sec(x) <-- Final Answer

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...

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...

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...