integration by substitution qn (1 Viewer)

Masaken · Jan 18, 2023

how to do this qn? i got stuck after doing the substitution u = 1 + sqrt(x), and i can't seem to get the substitution plugged in no matter what i do. help?

also, is there a general way on how to approach these sorts of questions where the substitution doesn't seem obvious to sub into initially? especially with square roots and exponentials, i have no idea

thanks in advance

yanujw · Jan 18, 2023

Let $u = \sqrt{x}$
$u^2 = x$
$dx = 2udu$
$2\int{\frac{u du}{1+u}}$
$2(\int 1 - \frac{1}{1+u}du)$
And complete via standard integrals and change back to x.

Determining substitutions is a skill you develop with practice. It is not always just to put the denominator as a variable, but try and imagine in your head what integral you get when performing a substitution before you write anything down to save time and effort.

Drongoski · Jan 18, 2023

Using my special method (just another way of doing substitution type integrals):

$\int \frac {1}{1 + \sqrt x} dx = \int \frac {1}{1 + \sqrt x} d(\sqrt x)^2 = \int \frac {1}{1 + \sqrt x} \times 2\sqrt x d \sqrt x \\ \\ = 2\int (1 - \frac {1}{1 + \sqrt x}) d\sqrt x = 2\int 1 d\sqrt x - 2\int \frac {1}{1 + \sqrt x} d(1+\sqrt x) \\ \\ = 2\sqrt x - 2ln (1 + \sqrt x) + C$

My solution above in much greater detail than is necessary to make it easier for most, unfamiliar with my weird method, to follow.

Note:
$dx^2 = 2xdx \cdots du^2 = 2udu \cdots d(\sqrt x)^2 = 2\sqrt x d \sqrt x \\ \\ d(x + 1 ) = dx \cdots d\sqrt x = d(1 + \sqrt x)$

integration by substitution qn (1 Viewer)

Masaken

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yanujw

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Drongoski

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