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Proofs of Selected Integrals

ò 1/(1+sqrt(ax))

Multiply the numerator and denominator of the fraction by sqrt(ax) giving

sqrt(ax) / ( (1+sqrt(ax)) (sqrt(ax)) )

And then express sqrt(ax) as 1+sqrt(ax) - 1, and split the fraction into two so our original integral becomes

dx/sqrt(ax) - dx/( (1+sqrt(ax)) (sqrt(ax)) )

Then the first integral is just (2/a) sqrt(ax), and the second can be solved using the substitution

u = 1 + sqrt(ax), so
du = (a/2) (dx / sqrt(ax)), and
(2/a) du = dx/sqrt(ax)

That way, our second integral becomes

(2/a) du/u

So the second integral is (2/a) ln(u) = (2/a) ln(1+sqrt(ax))

Related pages in this website

Table of Integrals

Integration by Parts

Partial Fractions


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