What is the integral from 0 to x^2 of 1/(1+t) ?

The integral from 0 to x^2 of 1/(1+t) is ln|x^2+1|

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integral from 0 to x^2 of 1/(1+t)

\int_{0}^{x^{2}} \frac{1}{1 + t} d t

ln x^{2} + 1

Solution steps

\int_{0}^{x^{2}} \frac{1}{1 + t} d t

Apply u-substitution

: \int_{1}^{1 + x^{2}} \frac{1}{u} d u

= \int_{1}^{1 + x^{2}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{1 + x^{2}}

Compute the boundaries:

ln x^{2} + 1

= ln x^{2} + 1

\int_{0}^{a} \frac{1}{a ^{2} + x ^{2}} d x

\int_{0}^{1} 5 x + 4 d x

\int_{0}^{x} 9 + x^{2} d x

\int_{0}^{5} (2 x + 1) d x

\int_{0}^{2} (x^{2} - 9 x + 14) d x

What is the integral from 0 to x^2 of 1/(1+t) ?
The integral from 0 to x^2 of 1/(1+t) is ln|x^2+1|