展開する-((x+1)^3(3ln(x+1)+1))/9

解

展開する

- \frac{( x + 1 ) ^{3} ( 3 ln ( x + 1 ) + 1 )}{9}

- \frac{x ^{3} ln ( x + 1 )}{3} - \frac{x ^{3}}{9} - x^{2} ln (x + 1) - \frac{x ^{2}}{3} - x ln (x + 1) - \frac{x}{3} - \frac{ln ( x + 1 )}{3} - \frac{1}{9}

解答ステップ

- \frac{( x + 1 ) ^{3} ( 3 ln ( x + 1 ) + 1 )}{9}

(x + 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1

= - \frac{( x ^{3} + 3 x ^{2} + 3 x + 1 ) ( 3 ln ( x + 1 ) + 1 )}{9}

拡張

(x^{3} + 3 x^{2} + 3 x + 1) (3 ln (x + 1) + 1) : 3 x^{3} ln (x + 1) + x^{3} + 9 x^{2} ln (x + 1) + 3 x^{2} + 9 x ln (x + 1) + 3 x + 3 ln (x + 1) + 1

= - \frac{3 x ^{3} ln ( x + 1 ) + x ^{3} + 9 x ^{2} ln ( x + 1 ) + 3 x ^{2} + 9 x ln ( x + 1 ) + 3 x + 3 ln ( x + 1 ) + 1}{9}

分数の規則を適用する:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{3 x ^{3} ln ( x + 1 ) + x ^{3} + 9 x ^{2} ln ( x + 1 ) + 3 x ^{2} + 9 x ln ( x + 1 ) + 3 x + 3 ln ( x + 1 ) + 1}{9} = - (\frac{3 x ^{3} ln ( x + 1 )}{9}) - (\frac{x ^{3}}{9}) - (\frac{9 x ^{2} ln ( x + 1 )}{9}) - (\frac{3 x ^{2}}{9}) - (\frac{9 x ln ( x + 1 )}{9}) - (\frac{3 x}{9}) - (\frac{3 ln ( x + 1 )}{9}) - (\frac{1}{9})

= - (\frac{3 x ^{3} ln ( x + 1 )}{9}) - (\frac{x ^{3}}{9}) - (\frac{9 x ^{2} ln ( x + 1 )}{9}) - (\frac{3 x ^{2}}{9}) - (\frac{9 x ln ( x + 1 )}{9}) - (\frac{3 x}{9}) - (\frac{3 ln ( x + 1 )}{9}) - (\frac{1}{9})

括弧を削除する:

(a) = a

= - \frac{3 x ^{3} ln ( x + 1 )}{9} - \frac{x ^{3}}{9} - \frac{9 x ^{2} ln ( x + 1 )}{9} - \frac{3 x ^{2}}{9} - \frac{9 x ln ( x + 1 )}{9} - \frac{3 x}{9} - \frac{3 ln ( x + 1 )}{9} - \frac{1}{9}

\frac{3 x ^{3} ln ( x + 1 )}{9} = \frac{x ^{3} ln ( x + 1 )}{3}

\frac{9 x ^{2} ln ( x + 1 )}{9} = x^{2} ln (x + 1)

\frac{3 x ^{2}}{9} = \frac{x ^{2}}{3}

\frac{9 x ln ( x + 1 )}{9} = x ln (x + 1)

\frac{3 x}{9} = \frac{x}{3}

\frac{3 ln ( x + 1 )}{9} = \frac{ln ( x + 1 )}{3}

= - \frac{x ^{3} ln ( x + 1 )}{3} - \frac{x ^{3}}{9} - x^{2} ln (x + 1) - \frac{x ^{2}}{3} - x ln (x + 1) - \frac{x}{3} - \frac{ln ( x + 1 )}{3} - \frac{1}{9}