int frac{2 t+1}{t^2+t+1} d t= | vedclass

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(A) To solve the integral $\int \frac{2 t+1}{t^2+t+1} d t$,we use the method of substitution.Let $u = t^2+t+1$.Then,the derivative of $u$ with respect

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$\log \left|t^2+t+1\right|+c$

Vedclass · Answer

(A) To solve the integral $\int \frac{2 t+1}{t^2+t+1} d t$,we use the method of substitution.Let $u = t^2+t+1$.Then,the derivative of $u$ with respect to $t$ is $\frac{du}{dt} = 2t+1$.This implies $du = (2t+1) dt$.Substituting these into the original integral,we get $\int \frac{1}{u} du$.The integral of $\frac{1}{u}$ is $\log |u| + c$.Substituting back $u = t^2+t+1$,we obtain $\log |t^2+t+1| + c$.

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

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