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

The integral from x to-2 of ln(1+t^2) is -2ln(5)-2arctan(2)+4-xln(1+x^2)-2arctan(x)+2x

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

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

- 2 ln (5) - 2 arctan (2) + 4 - x ln (1 + x^{2}) - 2 arctan (x) + 2 x

Solution steps

\int_{x}^{- 2} 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]_{x}^{- 2}

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

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

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

Compute the boundaries:

- 2 ln (5) - 2 arctan (2) + 4 - x ln (1 + x^{2}) - 2 arctan (x) + 2 x

= - 2 ln (5) - 2 arctan (2) + 4 - x ln (1 + x^{2}) - 2 arctan (x) + 2 x

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

\int_{0}^{1} (7 - 8 v^{3} + 16 v^{7}) d v

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

\int_{- 2}^{2} - 2 x^{2} d x

\int_{1}^{e^{5}} \frac{4 ln ( x )}{x} d x

What is the integral from x to-2 of ln(1+t^2) ?
The integral from x to-2 of ln(1+t^2) is -2ln(5)-2arctan(2)+4-xln(1+x^2)-2arctan(x)+2x