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

The integral from 1/e to tan(x) of t/(1+t^2) is 1/2 (ln(sec^2(x))-ln(1+1/(e^2)))

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integral from 1/e to tan(x) of t/(1+t^2)

\int_{\frac{1}{e}}^{t a n (x)} \frac{t}{1 + t ^{2}} d t

\frac{1}{2} (ln (sec^{2} (x)) - ln (1 + \frac{1}{e ^{2}}))

Solution steps

\int_{\frac{1}{e}}^{t a n (x)} \frac{t}{1 + t ^{2}} d t

Apply u-substitution

: \int_{1 + \frac{1}{e ^{2}}}^{1 + t a n^{2} (x)} \frac{1}{2 u} d u

= \int_{1 + \frac{1}{e ^{2}}}^{1 + t a n^{2} (x)} \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{1 + \frac{1}{e ^{2}}}^{1 + t a n^{2} (x)} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1 + \frac{1}{e ^{2}}}^{1 + t a n^{2} (x)}

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

= \frac{1}{2} [ln ∣ u ∣]_{1 + \frac{1}{e ^{2}}}^{s e c^{2} (x)}

Compute the boundaries:

ln (sec^{2} (x)) - ln (1 + \frac{1}{e ^{2}})

= \frac{1}{2} (ln (sec^{2} (x)) - ln (1 + \frac{1}{e ^{2}}))

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What is the integral from 1/e to tan(x) of t/(1+t^2) ?
The integral from 1/e to tan(x) of t/(1+t^2) is 1/2 (ln(sec^2(x))-ln(1+1/(e^2)))