What is the integral from 1 to e^{3x of}(ln(t))^2 ?

The integral from 1 to e^{3x of}(ln(t))^2 is 9x^2e^{3x}-2(3xe^{3x}-e^{3x})-2

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integral from 1 to e^{3x of}(ln(t))^2

\int_{1}^{e^{3 x}} (ln (t))^{2} d t

9 x^{2} e^{3 x} - 2 (3 x e^{3 x} - e^{3 x}) - 2

Solution steps

\int_{1}^{e^{3 x}} (ln (t))^{2} d t

Apply Integration By Parts

: [t ln^{2} (t) - \int 2 ln (t) d t]_{1}^{e^{3 x}}

= [t ln^{2} (t) - \int 2 ln (t) d t]_{1}^{e^{3 x}}

\int 2 ln (t) d t = 2 (t ln (t) - t)

= [t ln^{2} (t) - 2 (t ln (t) - t)]_{1}^{e^{3 x}}

Compute the boundaries:

9 x^{2} e^{3 x} - 2 (3 x e^{3 x} - e^{3 x}) - 2

= 9 x^{2} e^{3 x} - 2 (3 x e^{3 x} - e^{3 x}) - 2

\int_{0}^{4} (x^{3} + 5) d x

\int_{0}^{8.13} 1000 x + 250 d x

\int_{\frac{5}{3}}^{3} 3 t - 5 d t

\int_{5}^{11} \frac{1}{x - 4} d x

\int_{- 13}^{13} \frac{1}{x ^{2} + x - 56} d x

What is the integral from 1 to e^{3x of}(ln(t))^2 ?
The integral from 1 to e^{3x of}(ln(t))^2 is 9x^2e^{3x}-2(3xe^{3x}-e^{3x})-2