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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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