What is the integral of 2x(x^2+1)^{23} ?

The integral of 2x(x^2+1)^{23} is 1/24 (x^2+1)^{24}+C

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integral of 2x(x^2+1)^{23}

\int 2 x (x^{2} + 1)^{23} d x

\frac{1}{24} (x^{2} + 1)^{24} + C

Solution steps

\int 2 x (x^{2} + 1)^{23} d x

Take the constant out:

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

= 2 \cdot \int x (x^{2} + 1)^{23} d x

Apply u-substitution

: \int \frac{u ^{23}}{2} d u

= 2 \cdot \int \frac{u ^{23}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{23} d u

Apply the Power Rule

: \frac{u ^{24}}{24}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{24}}{24}

Substitute back

u = x^{2} + 1

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{24}}{24}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{24}}{24} : \frac{1}{24} (x^{2} + 1)^{24}

= \frac{1}{24} (x^{2} + 1)^{24}

Add a constant to the solution

= \frac{1}{24} (x^{2} + 1)^{24} + C

n = 1 \sum \infty - \frac{1}{3 ( 3 n + 2 )}

\int \frac{1}{y ln ( y )} d y

\frac{\partial}{\partial v} (u + 2 v)

\int_{1}^{2} \frac{v ^{6} + 4 v ^{8}}{v ^{5}} d v

y^{^{''}} + 4 y^{^{'}} - 16 y = 0

What is the integral of 2x(x^2+1)^{23} ?
The integral of 2x(x^2+1)^{23} is 1/24 (x^2+1)^{24}+C