What is the integral of (yln(y))/(1+x^2) ?

The integral of (yln(y))/(1+x^2) is ln(y)yarctan(x)+C

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integral of (yln(y))/(1+x^2)

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

ln (y) y arctan (x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= ln (y) y \cdot \int \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= ln (y) y arctan (x)

Add a constant to the solution

= ln (y) y arctan (x) + C

\int 2 x (x^{2} + 8)^{5} d x

\int arctan (2 x - 1) d x

\int \frac{10}{( x - 1 ) ^{2}} d x

\int \frac{1}{2 x - 9} d x

\int (e^{\frac{x}{2}})^{2} d x

What is the integral of (yln(y))/(1+x^2) ?
The integral of (yln(y))/(1+x^2) is ln(y)yarctan(x)+C