What is the integral of+(13)/((t^2+1)^{3/2)} ?

The integral of+(13)/((t^2+1)^{3/2)} is (13t)/((1+t^2)^{1/2)}+C

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integral of+(13)/((t^2+1)^{3/2)}

\int + \frac{13}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{13 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{13}{( 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

= 13 \cdot \int \frac{1}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

Apply Trigonometric Substitution

: \int cos (u) d u

= 13 \cdot \int cos (u) d u

Use the common integral:

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

= 13 sin (u)

Substitute back

u = arctan (t)

= 13 sin (arctan (t))

Simplify

13 sin (arctan (t)) : \frac{13 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

= \frac{13 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{13 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

x \to 1 lim (\frac{1}{x})

n \to \infty lim (tanh (n))

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

\int 4 e^{- t} t d t

\int sec (8 x) d x

What is the integral of+(13)/((t^2+1)^{3/2)} ?
The integral of+(13)/((t^2+1)^{3/2)} is (13t)/((1+t^2)^{1/2)}+C