What is expand x(u''(x)ln(x)+2u'(x)+\H(-1/(x^2))) ?

The solution to expand x(u''(x)ln(x)+2u'(x)+\H(-1/(x^2))) is xln(x)u''(x)+2xu'(x)+x\H(-1/(x^2))

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expand x(u''(x)ln(x)+2u'(x)+\H(-1/(x^2)))

expand

x (u^{''} (x) ln (x) + 2 u^{'} (x) + H (- \frac{1}{x ^{2}}))

x ln (x) u'' (x) + 2 xu' (x) + x H (- \frac{1}{x ^{2}})

Solution steps

x (u^{''} (x) ln (x) + 2 u^{'} (x) + H (- \frac{1}{x ^{2}}))

= x (ln (x) u^{^{''}} (x) + 2 u^{^{'}} (x) + H (- \frac{1}{x ^{2}}))

Distribute parentheses

= x u^{^{''}} (x) ln (x) + x \cdot 2 u^{^{'}} (x) + x H (- \frac{1}{x ^{2}})

= x ln (x) u^{^{''}} (x) + 2 x u^{^{'}} (x) + x H (- \frac{1}{x ^{2}})

3 x (x y - 2) d x

(e^{t} + e^{- t})^{2} t

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

(b ee)^{3} + b ee + 1

\frac{3 x + 5}{( x - 3 ) ( x + 4 )}

What is expand x(u''(x)ln(x)+2u'(x)+\H(-1/(x^2))) ?
The solution to expand x(u''(x)ln(x)+2u'(x)+\H(-1/(x^2))) is xln(x)u''(x)+2xu'(x)+x\H(-1/(x^2))