What is the integral of {(3-x)/(x^2+1)+xln(x^2)} ?

The integral of {(3-x)/(x^2+1)+xln(x^2)} is 3arctan(x)-1/2 ln|x^2+1|+x^2ln(x)-(x^2)/2+C

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integral of {(3-x)/(x^2+1)+xln(x^2)}

\int {\frac{3 - x}{x ^{2} + 1} + x ln (x^{2})} d x

3 arctan (x) - \frac{1}{2} ln x^{2} + 1 + x^{2} ln (x) - \frac{x ^{2}}{2} + C

Solution steps

\int \frac{3 - x}{x ^{2} + 1} + x ln (x^{2}) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3 - x}{x ^{2} + 1} d x + \int x ln (x^{2}) d x

\int \frac{3 - x}{x ^{2} + 1} d x = 3 arctan (x) - \frac{1}{2} ln x^{2} + 1

\int x ln (x^{2}) d x = x^{2} ln (x) - \frac{x ^{2}}{2}

= 3 arctan (x) - \frac{1}{2} ln x^{2} + 1 + x^{2} ln (x) - \frac{x ^{2}}{2}

Add a constant to the solution

= 3 arctan (x) - \frac{1}{2} ln x^{2} + 1 + x^{2} ln (x) - \frac{x ^{2}}{2} + C

y = 1 + 3 x^{2} (ln (x^{2}))

f (x) = (x^{3} + 4) e^{x}

\int \frac{x ^{2} + 4}{x + 1} d x

n = 0 \sum \infty 3 (\frac{x}{4})^{n}

f (x) = \frac{1}{x - 5}, at x = 10

What is the integral of {(3-x)/(x^2+1)+xln(x^2)} ?
The integral of {(3-x)/(x^2+1)+xln(x^2)} is 3arctan(x)-1/2 ln|x^2+1|+x^2ln(x)-(x^2)/2+C