well,
∫ lnx/x^2 dx = -(lnx + 1)/x
recall that ln √x = 1/2 lnx
let u = 3-5x^2 and du=-10x dx
That gives you
∫ -1/10 2^u du
which is just 2^u / ln u
Now it's pretty straightforward
Integrate and simplify the following function:
y= ∫(ln√x/x^2)+x 2^(3-5x^2))dx
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