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

\int sin^{4} (x co) s^{4} x d x

y = \frac{1}{20 x ^{2}}, y = x, y = \frac{x}{16}

\frac{\partial}{\partial y} (y^{\frac{3}{2}})

\int f (t) d t

\int \frac{( x - 2 )}{x ^{2} - 4 x + 9} d x

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