What is the integral of 4x^3+2xy^2ln(x^2+y^2) ?

The integral of 4x^3+2xy^2ln(x^2+y^2) is x^4-y^2x^2+y^2x^2ln(x^2+y^2)+y^4ln(x^2+y^2)-2y^4+C

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integral of 4x^3+2xy^2ln(x^2+y^2)

\int 4 x^{3} + 2 x y^{2} ln (x^{2} + y^{2}) d x

x^{4} - y^{2} x^{2} + y^{2} x^{2} ln (x^{2} + y^{2}) + y^{4} ln (x^{2} + y^{2}) - 2 y^{4} + C

Solution steps

\int 4 x^{3} + 2 x y^{2} ln (x^{2} + y^{2}) d x

Apply u-substitution

: \int 2 (u - y^{2}) + y^{2} ln (u) d u

= \int 2 (u - y^{2}) + y^{2} ln (u) d u

Apply the Sum Rule:

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

= \int 2 (u - y^{2}) d u + \int y^{2} ln (u) d u

\int 2 (u - y^{2}) d u = u^{2} - 2 y^{2} u

\int y^{2} ln (u) d u = y^{2} (u ln (u) - u)

= u^{2} - 2 y^{2} u + y^{2} (u ln (u) - u)

Substitute back

u = y^{2} + x^{2}

= (y^{2} + x^{2})^{2} - 2 y^{2} (y^{2} + x^{2}) + y^{2} ((y^{2} + x^{2}) ln (y^{2} + x^{2}) - (y^{2} + x^{2}))

Simplify

(y^{2} + x^{2})^{2} - 2 y^{2} (y^{2} + x^{2}) + y^{2} ((y^{2} + x^{2}) ln (y^{2} + x^{2}) - (y^{2} + x^{2})) : x^{4} - y^{2} x^{2} + y^{2} x^{2} ln (x^{2} + y^{2}) + y^{4} ln (x^{2} + y^{2}) - 2 y^{4}

= x^{4} - y^{2} x^{2} + y^{2} x^{2} ln (x^{2} + y^{2}) + y^{4} ln (x^{2} + y^{2}) - 2 y^{4}

Add a constant to the solution

= x^{4} - y^{2} x^{2} + y^{2} x^{2} ln (x^{2} + y^{2}) + y^{4} ln (x^{2} + y^{2}) - 2 y^{4} + C

(x^{2} - x y) d x + x d y = 0

\int \frac{x ^{3} + 2}{x ^{2} + 8 x + 7} d x

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

n = 0 \sum \infty \frac{7}{5 ^{n}}

\int - 7 sec (2 x) tan (2 x) d x

What is the integral of 4x^3+2xy^2ln(x^2+y^2) ?
The integral of 4x^3+2xy^2ln(x^2+y^2) is x^4-y^2x^2+y^2x^2ln(x^2+y^2)+y^4ln(x^2+y^2)-2y^4+C