What is the integral of (2t+1)/(t^2+t-4) ?

The integral of (2t+1)/(t^2+t-4) is ln|t^2+t-4|+C

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integral of (2t+1)/(t^2+t-4)

\int \frac{2 t + 1}{t ^{2} + t - 4} d t

ln t^{2} + t - 4 + C

Solution steps

\int \frac{2 t + 1}{t ^{2} + t - 4} d t

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = t^{2} + t - 4

= ln t^{2} + t - 4

Add a constant to the solution

= ln t^{2} + t - 4 + C

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What is the integral of (2t+1)/(t^2+t-4) ?
The integral of (2t+1)/(t^2+t-4) is ln|t^2+t-4|+C