What is the integral of 4/(tsqrt(t^2+36)) ?

The integral of 4/(tsqrt(t^2+36)) is 2/3 ln|tan(1/2 arctan(1/6 t))|+C

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integral of 4/(tsqrt(t^2+36))

\int \frac{4}{t t ^{2} + 36} d t

\frac{2}{3} ln tan (\frac{1}{2} arctan (\frac{1}{6} t)) + C

Solution steps

\int \frac{4}{t t ^{2} + 36} d t

Take the constant out:

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

= 4 \cdot \int \frac{1}{t t ^{2} + 36} d t

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{6 tan ( u )} d u

= 4 \cdot \int \frac{sec ( u )}{6 tan ( u )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{6} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

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

= 4 \cdot \frac{1}{6} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 4 \cdot \frac{1}{6} ln ∣ v ∣

Substitute back

= 4 \cdot \frac{1}{6} ln tan (\frac{arctan ( \frac{1}{6} t )}{2})

Simplify

4 \cdot \frac{1}{6} ln tan (\frac{arctan ( \frac{1}{6} t )}{2}) : \frac{2}{3} ln tan (\frac{1}{2} arctan (\frac{1}{6} t))

= \frac{2}{3} ln tan (\frac{1}{2} arctan (\frac{1}{6} t))

Add a constant to the solution

= \frac{2}{3} ln tan (\frac{1}{2} arctan (\frac{1}{6} t)) + C

\int x^{2} + 2 x^{4} d x

\frac{d}{d x} (\frac{sin ^{5} ( x )}{5} - \frac{sin ^{7} ( x )}{7})

\frac{d}{d x} (e^{x} - cos (x))

y = (2 x + 1) (x + 5)^{3}

n = 0 \sum \infty (- 1)^{n} (2^{n})

What is the integral of 4/(tsqrt(t^2+36)) ?
The integral of 4/(tsqrt(t^2+36)) is 2/3 ln|tan(1/2 arctan(1/6 t))|+C