What is the integral of x(1+x^2)^9 ?

The integral of x(1+x^2)^9 is 1/20 (1+x^2)^{10}+C

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integral of x(1+x^2)^9

\int x (1 + x^{2})^{9} d x

\frac{1}{20} (1 + x^{2})^{10} + C

Solution steps

\int x (1 + x^{2})^{9} d x

Apply u-substitution

: \int \frac{u ^{9}}{2} d u

= \int \frac{u ^{9}}{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^{9} d u

Apply the Power Rule

: \frac{u ^{10}}{10}

= \frac{1}{2} \cdot \frac{u ^{10}}{10}

Substitute back

u = 1 + x^{2}

= \frac{1}{2} \cdot \frac{( 1 + x ^{2} ) ^{10}}{10}

Simplify

\frac{1}{2} \cdot \frac{( 1 + x ^{2} ) ^{10}}{10} : \frac{1}{20} (1 + x^{2})^{10}

= \frac{1}{20} (1 + x^{2})^{10}

Add a constant to the solution

= \frac{1}{20} (1 + x^{2})^{10} + C

\int ln (10)

\int 3 x e^{\frac{1}{2} x^{2}} d x

\int x^{3} x^{3} d x

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

What is the integral of x(1+x^2)^9 ?
The integral of x(1+x^2)^9 is 1/20 (1+x^2)^{10}+C