What is the integral of (4(t^2+3))/(t(t^2+6)) ?

The integral of (4(t^2+3))/(t(t^2+6)) is ln|t^2+6|+2ln|t|+C

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integral of (4(t^2+3))/(t(t^2+6))

\int \frac{4 ( t ^{2} + 3 )}{t ( t ^{2} + 6 )} d t

ln t^{2} + 6 + 2 ln ∣ t ∣ + C

Solution steps

\int \frac{4 ( t ^{2} + 3 )}{t ( t ^{2} + 6 )} d t

Take the constant out:

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

= 4 \cdot \int \frac{t ^{2} + 3}{t ( t ^{2} + 6 )} d t

Expand

\frac{t ^{2} + 3}{t ( t ^{2} + 6 )} : \frac{t ^{2}}{t ( t ^{2} + 6 )} + \frac{3}{t ( t ^{2} + 6 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int \frac{t ^{2}}{t ( t ^{2} + 6 )} d t + \int \frac{3}{t ( t ^{2} + 6 )} d t)

\int \frac{t ^{2}}{t ( t ^{2} + 6 )} d t = \frac{1}{2} ln t^{2} + 6

\int \frac{3}{t ( t ^{2} + 6 )} d t = 3 (\frac{1}{6} ln ∣ t ∣ - \frac{1}{12} ln t^{2} + 6)

= 4 (\frac{1}{2} ln t^{2} + 6 + 3 (\frac{1}{6} ln ∣ t ∣ - \frac{1}{12} ln t^{2} + 6))

Simplify

4 (\frac{1}{2} ln t^{2} + 6 + 3 (\frac{1}{6} ln ∣ t ∣ - \frac{1}{12} ln t^{2} + 6)) : ln t^{2} + 6 + 2 ln ∣ t ∣

= ln t^{2} + 6 + 2 ln ∣ t ∣

Add a constant to the solution

= ln t^{2} + 6 + 2 ln ∣ t ∣ + C

\int \frac{1}{5 x ( ln ( x ) )} d x

\int \frac{4 x}{4 x + 1} d x

\int (2 x^{3} + 1)^{7} \cdot x^{2} d x

\int e^{- \frac{3}{x}} d x

\int (2 x + 5 y) d y

What is the integral of (4(t^2+3))/(t(t^2+6)) ?
The integral of (4(t^2+3))/(t(t^2+6)) is ln|t^2+6|+2ln|t|+C