What is the integral of 1/(t[1+(ln(t))^2]) ?

The integral of 1/(t[1+(ln(t))^2]) is arctan(ln(t))+C

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integral of 1/(t[1+(ln(t))^2])

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

arctan (ln (t)) + C

Solution steps

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

Simplify

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

Apply u-substitution

: \int \frac{1}{1 + u ^{2}} d u

= \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= arctan (u)

Substitute back

u = ln (t)

= arctan (ln (t))

Add a constant to the solution

= arctan (ln (t)) + C

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What is the integral of 1/(t[1+(ln(t))^2]) ?
The integral of 1/(t[1+(ln(t))^2]) is arctan(ln(t))+C