What is the integral from x to 8 of 1/(sqrt(1+t^2)) ?

The integral from x to 8 of 1/(sqrt(1+t^2)) is ln(sqrt(65)+8)-ln|sqrt(x^2+1)+x|

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integral from x to 8 of 1/(sqrt(1+t^2))

\int_{x}^{8} \frac{1}{1 + t ^{2}} d t

ln (65 + 8) - ln x^{2} + 1 + x

Solution steps

\int_{x}^{8} \frac{1}{1 + t ^{2}} d t

Use the common integral:

\int \frac{1}{1 + t ^{2}} d t = ln t^{2} + 1 + t

= [ln t^{2} + 1 + t]_{x}^{8}

Compute the boundaries:

ln (65 + 8) - ln x^{2} + 1 + x

= ln (65 + 8) - ln x^{2} + 1 + x

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What is the integral from x to 8 of 1/(sqrt(1+t^2)) ?
The integral from x to 8 of 1/(sqrt(1+t^2)) is ln(sqrt(65)+8)-ln|sqrt(x^2+1)+x|