2 years ago

1 Calculate:

1 answer:

5 0

(a)=940
(b)=1680
(c)=1950
(d)=4440

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photoshop1234 [79]

Substitute $u=x^2+4$ and $du=2x\,dx$ . Then the integral transforms to

$\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \int \frac{du}{u^3}$

Apply the power rule.

$\displaystyle \int \frac{du}{u^3} = -\dfrac1{2u^2} + C$

Now put the result back in terms of $x$ .

$\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \left(-\dfrac1{2u^2} + C\right) = -\dfrac1{4u^2} + C = \boxed{-\dfrac1{4(x^2+4)^2} + C}$

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