What is the integral of ((2t+1)^2)/(3t) ?

The integral of ((2t+1)^2)/(3t) is 1/3 (2t^2+4t+ln|t|+1)+C

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integral of ((2t+1)^2)/(3t)

\int \frac{( 2 t + 1 ) ^{2}}{3 t} d t

\frac{1}{3} (2 t^{2} + 4 t + ln ∣ t ∣ + 1) + C

Solution steps

\int \frac{( 2 t + 1 ) ^{2}}{3 t} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{( 2 t + 1 ) ^{2}}{t} d t

Apply Integration By Parts

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

= \frac{1}{3} ((2 t + 1)^{2} - \int \frac{4 t ^{2} - 1}{t} d t)

\int \frac{4 t ^{2} - 1}{t} d t = 2 t^{2} - ln ∣ t ∣

= \frac{1}{3} ((2 t + 1)^{2} - (2 t^{2} - ln ∣ t ∣))

Simplify

\frac{1}{3} ((2 t + 1)^{2} - (2 t^{2} - ln ∣ t ∣)) : \frac{1}{3} (2 t^{2} + 4 t + ln ∣ t ∣ + 1)

= \frac{1}{3} (2 t^{2} + 4 t + ln ∣ t ∣ + 1)

Add a constant to the solution

= \frac{1}{3} (2 t^{2} + 4 t + ln ∣ t ∣ + 1) + C

\int (3 x + a) sin (x) d x

\int \frac{5 x ^{3} - 3 x ^{2} + 7 x - 3}{( x ^{2} + 1 ) ^{2}} d x

\int x (2 x) d x

\int x_{1} d x

What is the integral of ((2t+1)^2)/(3t) ?
The integral of ((2t+1)^2)/(3t) is 1/3 (2t^2+4t+ln|t|+1)+C