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人気のある微分積分 >

integral of tln(sqrt(t+1))

解

\int t ln (t + 1) d t

解

\frac{1}{8} (2 t^{2} ln (t + 1) - t^{2} + 2 t - 2 ln (t + 1) + 3) + C

解答ステップ

\int t ln (t + 1) d t

簡素化

ln (t + 1) : \frac{1}{2} ln (t + 1)

= \int \frac{1}{2} t ln (t + 1) d t

定数を除く:

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

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

u置換積分法を適用する

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

部分積分を適用する

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

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= \frac{1}{2} (ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u))

代用を戻す

u = t + 1

= \frac{1}{2} (ln (t + 1) (\frac{( t + 1 ) ^{2}}{2} - (t + 1)) - (\frac{( t + 1 ) ^{2}}{4} - (t + 1)))

簡素化

\frac{1}{2} (ln (t + 1) (\frac{( t + 1 ) ^{2}}{2} - (t + 1)) - (\frac{( t + 1 ) ^{2}}{4} - (t + 1))) : \frac{1}{8} (2 t^{2} ln (t + 1) - t^{2} + 2 t - 2 ln (t + 1) + 3)

= \frac{1}{8} (2 t^{2} ln (t + 1) - t^{2} + 2 t - 2 ln (t + 1) + 3)

定数を解答に追加する

= \frac{1}{8} (2 t^{2} ln (t + 1) - t^{2} + 2 t - 2 ln (t + 1) + 3) + C

グラフ

人気の例

integral of cos(2x-y)

\int cos (2 x - y) d x

integral of ((5x^2+3x+5))/((x^2+1)^2)

\int \frac{( 5 x ^{2} + 3 x + 5 )}{( x ^{2} + 1 ) ^{2}} d x

integral of ((x-1)^2)/(x+1)

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

integral of (xln(x))/(sqrt(x^2-4))

\int \frac{x ln ( x )}{x ^{2} - 4} d x

integral of (12x^3)/((3x^4-4)ln(3x^4-4))

\int \frac{12 x ^{3}}{( 3 x ^{4} - 4 ) ln ( 3 x ^{4} - 4 )} d x