Question

с 1м и 2м интегралом (35б)

Answer 1

Пошаговое объяснение:

1)

$\int\limits {(3x+1)^{\frac{2}{3} }} \, dx =\frac{1}{3}\int\limits {t}^{\frac{2}{3} } \, dt=\\=\frac{1}{3}*\frac{t^{\frac{2}{3} +1}}{\frac{2}{3} +1}= \frac{1}{3}* \frac{3}{5}t^{\frac{5}{3}}=\frac{1}{5}*(3x+1)^{\frac{5}{3}}\\3x+1=t; 3dx=dt; \\dx=\frac{1}{3}dt$

2)$\int \frac{dx}{xln^2x}=\int \frac{d(lnx)}{ln^2x}=\int \frac{dt}{t^2}=-\frac{1}{t}=-\frac{1}{lnx}$

Answer 2

Пошаговое объяснение:

1)

\begin{gathered}\int\limits {(3x+1)^{\frac{2}{3} }} \, dx =\frac{1}{3}\int\limits {t}^{\frac{2}{3} } \, dt=\\=\frac{1}{3}*\frac{t^{\frac{2}{3} +1}}{\frac{2}{3} +1}= \frac{1}{3}* \frac{3}{5}t^{\frac{5}{3}}=\frac{1}{5}*(3x+1)^{\frac{5}{3}}\\3x+1=t; 3dx=dt; \\dx=\frac{1}{3}dt\end{gathered}

∫(3x+1)

3

2

dx=

3

1

∫t

3

2

dt=

=

3

1

∗

3

2

+1

t

3

2

+1

=

3

1

∗

5

3

t

3

5

=

5

1

∗(3x+1)

3

5

3x+1=t;3dx=dt;

dx=

3

1

dt

2)\int \frac{dx}{xln^2x}=\int \frac{d(lnx)}{ln^2x}=\int \frac{dt}{t^2}=-\frac{1}{t}=-\frac{1}{lnx}∫

xln

2

x

dx

=∫

ln

2

x

d(lnx)

=∫

t

2

dt

=−

t

1

=−

lnx

1