What is the integral of ln(1+r^2)r ?

The integral of ln(1+r^2)r is 1/2 (ln(1+r^2)+r^2ln(1+r^2)-1-r^2)+C

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integral of ln(1+r^2)r

\int ln (1 + r^{2}) r d r

\frac{1}{2} (ln (1 + r^{2}) + r^{2} ln (1 + r^{2}) - 1 - r^{2}) + C

Solution steps

\int ln (1 + r^{2}) r d r

Apply u-substitution

: \int \frac{ln ( u )}{2} d u

= \int \frac{ln ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2} (u ln (u) - u)

Substitute back

u = 1 + r^{2}

= \frac{1}{2} ((1 + r^{2}) ln (1 + r^{2}) - (1 + r^{2}))

Expand

(1 + r^{2}) ln (1 + r^{2}) - (1 + r^{2}) : ln (1 + r^{2}) + r^{2} ln (1 + r^{2}) - 1 - r^{2}

= \frac{1}{2} (ln (1 + r^{2}) + r^{2} ln (1 + r^{2}) - 1 - r^{2})

Add a constant to the solution

= \frac{1}{2} (ln (1 + r^{2}) + r^{2} ln (1 + r^{2}) - 1 - r^{2}) + C

\frac{\partial}{\partial x} (yz ln (x y))

y (x + y + 1) d x + (x + 2 y) d y = 0

x^{^{''}} - 2 x - 5 = 0, x (0) = 1, x (1) = 3

\int \frac{39}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

5 x (1 - \frac{x}{5})^{2}

What is the integral of ln(1+r^2)r ?
The integral of ln(1+r^2)r is 1/2 (ln(1+r^2)+r^2ln(1+r^2)-1-r^2)+C