What is the integral of x(x^2+1)^{100} ?

The integral of x(x^2+1)^{100} is 1/202 (x^2+1)^{101}+C

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integral of x(x^2+1)^{100}

\int x (x^{2} + 1)^{100} d x

\frac{1}{202} (x^{2} + 1)^{101} + C

Solution steps

\int x (x^{2} + 1)^{100} d x

Apply u-substitution

: \int \frac{u ^{100}}{2} d u

= \int \frac{u ^{100}}{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^{100} d u

Apply the Power Rule

: \frac{u ^{101}}{101}

= \frac{1}{2} \cdot \frac{u ^{101}}{101}

Substitute back

u = x^{2} + 1

= \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{101}}{101}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{101}}{101} : \frac{1}{202} (x^{2} + 1)^{101}

= \frac{1}{202} (x^{2} + 1)^{101}

Add a constant to the solution

= \frac{1}{202} (x^{2} + 1)^{101} + C

\frac{x ^{2} - 10 x + 65}{x - 8}

n = 0 \sum \infty \frac{1}{( 2 n - 1 ) ^{2}}

\frac{d}{d x} (ln (cos (9 x)))

x \to 5 lim (\frac{sin ( 5 π )}{3})

n = 1 \sum \infty \frac{40}{4 ^{n}}

What is the integral of x(x^2+1)^{100} ?
The integral of x(x^2+1)^{100} is 1/202 (x^2+1)^{101}+C