What is the integral of (x^{15})/(sqrt(x^{16)+1)} ?

The integral of (x^{15})/(sqrt(x^{16)+1)} is 1/8 sqrt(x^{16)+1}+C

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integral of (x^{15})/(sqrt(x^{16)+1)}

\int \frac{x ^{15}}{x ^{16} + 1} d x

\frac{1}{8} x^{16} + 1 + C

Solution steps

\int \frac{x ^{15}}{x ^{16} + 1} d x

Apply u-substitution

: \int \frac{1}{16 u} d u

= \int \frac{1}{16 u} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{16} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{16} + 1

= \frac{1}{16} \cdot 2 (x^{16} + 1)^{\frac{1}{2}}

Simplify

\frac{1}{16} \cdot 2 (x^{16} + 1)^{\frac{1}{2}} : \frac{1}{8} x^{16} + 1

= \frac{1}{8} x^{16} + 1

Add a constant to the solution

= \frac{1}{8} x^{16} + 1 + C

\int 2 x (x^{2} + 2)^{7} d x

\frac{d}{d x} (- 2 x + 1)

10 x^{2}

\frac{d}{d x} (\frac{2 x ^{2} - 3}{x ^{2} + 1})

(5 x^{2})^{^{'}}

What is the integral of (x^{15})/(sqrt(x^{16)+1)} ?
The integral of (x^{15})/(sqrt(x^{16)+1)} is 1/8 sqrt(x^{16)+1}+C