What is the integral of t/((t^2+1)^{23/2)} ?

The integral of t/((t^2+1)^{23/2)} is -1/(21(t^2+1)^{21/2)}+C

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integral of t/((t^2+1)^{23/2)}

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{23}{2}}} d t

- \frac{1}{21 ( t ^{2} + 1 ) ^{\frac{21}{2}}} + C

Solution steps

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{23}{2}}} d t

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{23}{2}}} d u

= \int \frac{1}{2 u ^{\frac{23}{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{23}{2}}} d u

Apply the Power Rule

: - \frac{2}{21 u ^{\frac{21}{2}}}

= \frac{1}{2} (- \frac{2}{21 u ^{\frac{21}{2}}})

Substitute back

u = t^{2} + 1

= \frac{1}{2} (- \frac{2}{21 ( t ^{2} + 1 ) ^{\frac{21}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{21 ( t ^{2} + 1 ) ^{\frac{21}{2}}}) : - \frac{1}{21 ( t ^{2} + 1 ) ^{\frac{21}{2}}}

= - \frac{1}{21 ( t ^{2} + 1 ) ^{\frac{21}{2}}}

Add a constant to the solution

= - \frac{1}{21 ( t ^{2} + 1 ) ^{\frac{21}{2}}} + C

\int 6 ln (x) d x

2 x^{3} + 9 x^{2} + 9 x + 6

n = 0 \sum \infty \frac{1}{( 0.5 ) ^{n}}

x = 2 y, x = y^{2} - 15

x \to 0 lim (\frac{( 3 + x ) ^{3} - 9}{x})

What is the integral of t/((t^2+1)^{23/2)} ?
The integral of t/((t^2+1)^{23/2)} is -1/(21(t^2+1)^{21/2)}+C