What is the integral from 1 to 6 of 1/(sqrt(1+x^2)) ?

The integral from 1 to 6 of 1/(sqrt(1+x^2)) is ln(sqrt(37)+6)-ln(sqrt(2)+1)

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integral from 1 to 6 of 1/(sqrt(1+x^2))

\int_{1}^{6} \frac{1}{1 + x ^{2}} d x

ln (37 + 6) - ln (2 + 1)

Decimal Notation

1.61040 \dots

Solution steps

\int_{1}^{6} \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= [ln x^{2} + 1 + x]_{1}^{6}

Compute the boundaries:

ln (37 + 6) - ln (2 + 1)

= ln (37 + 6) - ln (2 + 1)

\int_{- 1}^{2} 2 x^{2} - 4 x + 5 d x

\int_{0}^{8} \frac{1}{v ^{\frac{1}{3}} + 1} d v

\int_{0}^{2} π (x^{2} - 4)^{2} d x

\int_{0}^{2 π} 9 d x

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

What is the integral from 1 to 6 of 1/(sqrt(1+x^2)) ?
The integral from 1 to 6 of 1/(sqrt(1+x^2)) is ln(sqrt(37)+6)-ln(sqrt(2)+1)