- #1
karush
Gold Member
MHB
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- 5
W8.5.7
Evaluate
$$\displaystyle \int\frac{{x}^{3}}{{x}^{2 }+4}\ dx $$
The book answer was
$$\displaystyle\frac{{x}^{2}}{2 }
-2\ln\left({4+{x}^{2 }}\right)+C$$
So
$$\displaystyle u=x^2 +4\ \ \ \ du=2x\ \ dx \ \ x=\sqrt{u-4} $$
Then
$$\displaystyle \int\frac{{x}^{2}}{{x}^{2 }+4}x\ dx
\implies \frac{1}{2} \int \frac{u-4}{u}\ \ du$$
I continued but didn't get the answer
Evaluate
$$\displaystyle \int\frac{{x}^{3}}{{x}^{2 }+4}\ dx $$
The book answer was
$$\displaystyle\frac{{x}^{2}}{2 }
-2\ln\left({4+{x}^{2 }}\right)+C$$
So
$$\displaystyle u=x^2 +4\ \ \ \ du=2x\ \ dx \ \ x=\sqrt{u-4} $$
Then
$$\displaystyle \int\frac{{x}^{2}}{{x}^{2 }+4}x\ dx
\implies \frac{1}{2} \int \frac{u-4}{u}\ \ du$$
I continued but didn't get the answer
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