How do you integrate by substitution $\int \frac{1}{2 \sqrt{x}} d x$ $int 1/(2sqrt(x))dx$ ?

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How do you integrate by substitution $\int \frac{1}{2 \sqrt{x}} d x$ $int 1/(2sqrt(x))dx$ ?

1 Answer

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Answer 1 · 2016-11-18T13:34:24

There is no need to use substitution.

Explanation:

$\int \frac{1}{2 \sqrt{x}} d x = \frac{1}{2} \int x^{- (\frac{1}{2})} d x$ $int 1/(2sqrtx) dx = 1/2 int x^-(1/2) dx$

Now use the power rule for integrals.

$= \frac{1}{2} 2 \cdot x^{\frac{1}{2}} + C$ $= 1/2 2*x^(1/2) +C$

$= x^{\frac{1}{2}} + C$ $= x^(1/2) +C$

IF someone insisted that I must do a substitution, I would let $u = x$ $u = x$ so $d u = d x$ $du = dx$ and integrate $\frac{1}{2 \sqrt{u}} d u$ $1/(2sqrtu) du$ exactly as above.

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How do you integrate by substitution $\int \frac{1}{2 \sqrt{x}} d x$ $int 1/(2sqrt(x))dx$ ?

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