What is the integral of (10x)/(81x^4+49) ?

The integral of (10x)/(81x^4+49) is 5/63 arctan(9/7 x^2)+C

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integral of (10x)/(81x^4+49)

\int \frac{10 x}{81 x ^{4} + 49} d x

\frac{5}{63} arctan (\frac{9}{7} x^{2}) + C

Solution steps

\int \frac{10 x}{81 x ^{4} + 49} d x

Take the constant out:

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

= 10 \cdot \int \frac{x}{81 x ^{4} + 49} d x

Apply u-substitution

: \int \frac{1}{2 ( 81 u ^{2} + 49 )} d u

= 10 \cdot \int \frac{1}{2 ( 81 u ^{2} + 49 )} d u

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \int \frac{1}{81 u ^{2} + 49} d u

Apply Integral Substitution

= 10 \cdot \frac{1}{2} \cdot \int \frac{1}{63 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \frac{1}{63} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 10 \cdot \frac{1}{2} \cdot \frac{1}{63} arctan (v)

Substitute back

= 10 \cdot \frac{1}{2} \cdot \frac{1}{63} arctan (\frac{9}{7} x^{2})

Simplify

10 \cdot \frac{1}{2} \cdot \frac{1}{63} arctan (\frac{9}{7} x^{2}) : \frac{5}{63} arctan (\frac{9}{7} x^{2})

= \frac{5}{63} arctan (\frac{9}{7} x^{2})

Add a constant to the solution

= \frac{5}{63} arctan (\frac{9}{7} x^{2}) + C

\int arccos (x) d x

\int_{- 2}^{2} (x^{3} - 1) d x

\frac{\partial}{\partial x} (x^{3} + x^{2} y^{3} - 2 y^{2})

f (x) = (6 x^{2} - 9 x) e^{x}

What is the integral of (10x)/(81x^4+49) ?
The integral of (10x)/(81x^4+49) is 5/63 arctan(9/7 x^2)+C