What is the integral of (14)/((t^2+1)^{3/2)} ?

The integral of (14)/((t^2+1)^{3/2)} is (14t)/((1+t^2)^{1/2)}+C

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integral of (14)/((t^2+1)^{3/2)}

\int \frac{14}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{14 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{14}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

Take the constant out:

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

= 14 \cdot \int \frac{1}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

Apply Trigonometric Substitution

: \int cos (u) d u

= 14 \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= 14 sin (u)

Substitute back

u = arctan (t)

= 14 sin (arctan (t))

Simplify

14 sin (arctan (t)) : \frac{14 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

= \frac{14 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{14 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

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

\frac{\partial}{\partial x} (\frac{( 2 x - 4 y )}{2 x + 4 y})

\int sin^{- 2} (x) d x

\frac{d}{d x} (20 - x^{2})

x \to 6 lim (f (x))

What is the integral of (14)/((t^2+1)^{3/2)} ?
The integral of (14)/((t^2+1)^{3/2)} is (14t)/((1+t^2)^{1/2)}+C