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

The integral from 1 to x^2 of ln(t) is 2x^2ln(x)-x^2+1

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integral from 1 to x^2 of ln(t)

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

2 x^{2} ln (x) - x^{2} + 1

Solution steps

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

Apply Integration By Parts

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

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

\int 1 d t = t

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

Compute the boundaries:

2 x^{2} ln (x) - x^{2} + 1

= 2 x^{2} ln (x) - x^{2} + 1

\int_{- 1}^{1} 3 x^{2} e^{3 x^{3} - 7} d x

\int_{0}^{2} 11 x (x^{2} + 1)^{2} d x

\int_{1}^{2} (2 x^{3} - 1)^{2} 6 x^{2} d x

\int_{\frac{π}{6}}^{\frac{π}{2}} sin^{2} (x) d x

\int_{- 2}^{4} (- \frac{x ^{2}}{2} + x + 4) d x

What is the integral from 1 to x^2 of ln(t) ?
The integral from 1 to x^2 of ln(t) is 2x^2ln(x)-x^2+1