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

The integral of t/((1+t^2)) is 1/2 ln|1+t^2|+C

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integral of t/((1+t^2))

integral

\frac{t}{( 1 + t ^{2} )}

\frac{1}{2} ln 1 + t^{2} + C

Solution steps

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

Apply u-substitution

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

= \int \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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + t^{2}

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

Add a constant to the solution

= \frac{1}{2} ln 1 + t^{2} + C

(1 - x^{2})^{\frac{1}{2}}

x (2 - x)^{3}

\frac{x}{( x ^{2} + 1 ) ^{2}}

\frac{1}{sin ^{4} ( x )}

\frac{tan ( x )}{cos ^{2} ( x )}

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