What is the integral of x(1+x^2)^{15} ?

The integral of x(1+x^2)^{15} is 1/32 (1+x^2)^{16}+C

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integral of x(1+x^2)^{15}

integral

x (1 + x^{2})^{15}

\frac{1}{32} (1 + x^{2})^{16} + C

Solution steps

\int x (1 + x^{2})^{15} d x

Apply u-substitution

: \int \frac{u ^{15}}{2} d u

= \int \frac{u ^{15}}{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 u^{15} d u

Apply the Power Rule

: \frac{u ^{16}}{16}

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

Substitute back

u = 1 + x^{2}

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

Simplify

\frac{1}{2} \cdot \frac{( 1 + x ^{2} ) ^{16}}{16} : \frac{1}{32} (1 + x^{2})^{16}

= \frac{1}{32} (1 + x^{2})^{16}

Add a constant to the solution

= \frac{1}{32} (1 + x^{2})^{16} + C

\frac{x}{x ^{2} + y ^{2}}

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

\frac{1}{4 - x ^{2} - 4 x}

\frac{1}{ln ( y )}

- t^{2} + 2

What is the integral of x(1+x^2)^{15} ?
The integral of x(1+x^2)^{15} is 1/32 (1+x^2)^{16}+C