What is the integral from 0 to x^2 of ln(t^2+1) ?

The integral from 0 to x^2 of ln(t^2+1) is x^2ln(x^4+1)-2(x^2-arctan(x^2))

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integral from 0 to x^2 of ln(t^2+1)

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

x^{2} ln (x^{4} + 1) - 2 (x^{2} - arctan (x^{2}))

Solution steps

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

Apply Integration By Parts

: [t ln (t^{2} + 1) - \int \frac{2 t ^{2}}{t ^{2} + 1} d t]_{0}^{x^{2}}

= [t ln (t^{2} + 1) - \int \frac{2 t ^{2}}{t ^{2} + 1} d t]_{0}^{x^{2}}

\int \frac{2 t ^{2}}{t ^{2} + 1} d t = 2 (- arctan (t) + t)

= [t ln (t^{2} + 1) - 2 (- arctan (t) + t)]_{0}^{x^{2}}

Compute the boundaries:

x^{2} ln (x^{4} + 1) - 2 (x^{2} - arctan (x^{2}))

= x^{2} ln (x^{4} + 1) - 2 (x^{2} - arctan (x^{2}))

\int_{4}^{9} \frac{1}{x - x} d x

\int_{3}^{5} \frac{1}{( x - 3 ) ^{\frac{3}{2}}} d x

\int_{- \infty}^{6} e^{x} d x

\int_{0}^{1} \frac{x ^{2} - 5}{1 + x ^{2}} d x

\int_{1}^{x + c} 1 d x

What is the integral from 0 to x^2 of ln(t^2+1) ?
The integral from 0 to x^2 of ln(t^2+1) is x^2ln(x^4+1)-2(x^2-arctan(x^2))