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integral ( (x^2+3x+7)/(sqrt(x)) ) dx?

how do u do that by sustitution method??

Answers (3)

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aa10667595 profile image
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8b24df9325c9 profile image

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

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qg2b7vmmaa profile image

( 2x^(5/2) )/5 + 2x^(3/2) + 14√x + C

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