integral of xln(x^2+1)+(x^2+1)y

Lösung

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

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

Schritte zur Lösung

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

Wende die Summenregel an:

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{1}{5 - x ^{2}} d x

y^{^{''}} + 16 y = 5 csc^{2} (4 t)

\int_{0}^{1} π (1 - x^{3})^{2} d x

\frac{\partial}{\partial x} (e^{x} \cdot sin (y))

\int_{0}^{\frac{1}{2}} (4 x + 3) e^{2 x} d x