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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

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

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

\int_{3}^{7} 10 x d x

\int_{1}^{2} \frac{1}{t ^{2}} d t

\int_{1}^{π + 1} sin (2 x) d x

\int_{0}^{4} 1 + x^{2} d x

\int_{3}^{5} \frac{1}{x} d x

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