how do u do that by sustitution method??

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how do u do that by sustitution method??

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## Answers (3)

Divide throughout by x^1/2 to get

x^3/2 + 3x^1/2 + 7x^-1/2

the integral of which is

(2/5) *x^(5/2) + 2*x^(3/2) + 14*x^(1/2) + constant

If you wish to simplify it to look like your original expression, you could "take out" an x^1/2:

sqrt(x) * [(2/5)*x^2 + 2*x + 14]

= 2* sqrt(x) / 5 * (x^2 + 5x + 35)

There's no dire need to use substitution.

Integral ( (x^2 + 3x + 7)/sqrt(x) dx )

sqrt(x) = x^(1/2), so we have

Integral ( (x^2 + 3x + 7)/x^(1/2) dx )

Split into three fractions.

Integral ( [ x^2/x^(1/2) + 3x/x^(1/2) + 7/x^(1/2) ] dx )

We can reduce these such that they are in the form ax^n.

Integral ( [ x^(3/2) + 3x^(1/2) + 7x^(-1/2) ] dx )

And now, use the reverse power rule. Remember that the integral of ax^n is equal to (a/(n + 1)) x^(n + 1)

(2/5)x^(5/2) + (3)(2/3)x^(3/2) + 7(2)x^(1/2) + C

Which simplifies as

(2/5)x^(5/2) + x^(3/2) + 14x^(1/2) + C

( 2x^(5/2) )/5 + 2x^(3/2) + 14âx + C

You don't need substitution, just divide through the âx and integrate each term.