What is the integral of (1+3t^5)/(2t) ?

The integral of (1+3t^5)/(2t) is 1/2 (ln|t|+(3t^5)/5)+C

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integral of (1+3t^5)/(2t)

\int \frac{1 + 3 t ^{5}}{2 t} d t

\frac{1}{2} (ln ∣ t ∣ + \frac{3 t ^{5}}{5}) + C

Solution steps

\int \frac{1 + 3 t ^{5}}{2 t} d t

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 + 3 t ^{5}}{t} d t

Expand

\frac{1 + 3 t ^{5}}{t} : \frac{1}{t} + 3 t^{4}

= \frac{1}{2} \cdot \int \frac{1}{t} + 3 t^{4} d t

Apply the Sum Rule:

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

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

\int \frac{1}{t} d t = ln ∣ t ∣

\int 3 t^{4} d t = \frac{3 t ^{5}}{5}

= \frac{1}{2} (ln ∣ t ∣ + \frac{3 t ^{5}}{5})

Add a constant to the solution

= \frac{1}{2} (ln ∣ t ∣ + \frac{3 t ^{5}}{5}) + C

\int (\frac{2 x ^{4}}{3 x ^{2}}) d x

y^{^{''}} + 36 y = - 36 sec^{2} (6 x)

f (x) = x + 6

x = 7 y, \frac{1}{3} y = x, \frac{1}{7} y = x

\frac{d}{d x} (ln (1 + sin^{2} (x)))

What is the integral of (1+3t^5)/(2t) ?
The integral of (1+3t^5)/(2t) is 1/2 (ln|t|+(3t^5)/5)+C