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

The integral of t/((t^2+1)^{13/2)} is -1/(11(t^2+1)^{11/2)}+C

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

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

- \frac{1}{11 ( t ^{2} + 1 ) ^{\frac{11}{2}}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{13}{2}}} d u

= \int \frac{1}{2 u ^{\frac{13}{2}}} d u

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}{u ^{\frac{13}{2}}} d u

Apply the Power Rule

: - \frac{2}{11 u ^{\frac{11}{2}}}

= \frac{1}{2} (- \frac{2}{11 u ^{\frac{11}{2}}})

Substitute back

u = t^{2} + 1

= \frac{1}{2} (- \frac{2}{11 ( t ^{2} + 1 ) ^{\frac{11}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{11 ( t ^{2} + 1 ) ^{\frac{11}{2}}}) : - \frac{1}{11 ( t ^{2} + 1 ) ^{\frac{11}{2}}}

= - \frac{1}{11 ( t ^{2} + 1 ) ^{\frac{11}{2}}}

Add a constant to the solution

= - \frac{1}{11 ( t ^{2} + 1 ) ^{\frac{11}{2}}} + C

x \to 0 lim (\frac{tan ( 2 x )}{x})

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

x^{^{'}} = 2 - x, x (5) = 4

x \to 0 lim ((x^{4}) cos (\frac{2}{x}))

x y^{^{'}} = (1 - x) y + x^{2} + x^{3}, y (3) = 2

What is the integral of t/((t^2+1)^{13/2)} ?
The integral of t/((t^2+1)^{13/2)} is -1/(11(t^2+1)^{11/2)}+C