What is the integral of (4u)/(16u^2+7) ?

The integral of (4u)/(16u^2+7) is 1/8 ln|16u^2+7|+C

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integral of (4u)/(16u^2+7)

\int \frac{4 u}{16 u ^{2} + 7} d u

\frac{1}{8} ln 16 u^{2} + 7 + C

Solution steps

\int \frac{4 u}{16 u ^{2} + 7} d u

Take the constant out:

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

= 4 \cdot \int \frac{u}{16 u ^{2} + 7} d u

Apply u-substitution

: \int \frac{1}{32 v} d v

= 4 \cdot \int \frac{1}{32 v} d v

Take the constant out:

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

= 4 \cdot \frac{1}{32} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 4 \cdot \frac{1}{32} ln ∣ v ∣

Substitute back

v = 16 u^{2} + 7

= 4 \cdot \frac{1}{32} ln 16 u^{2} + 7

Simplify

4 \cdot \frac{1}{32} ln 16 u^{2} + 7 : \frac{1}{8} ln 16 u^{2} + 7

= \frac{1}{8} ln 16 u^{2} + 7

Add a constant to the solution

= \frac{1}{8} ln 16 u^{2} + 7 + C

\frac{d}{d x} (e^{9 x s i n (2 x)})

n = 1 \sum \infty \frac{cos ( n )}{n ^{5}}

\frac{d}{d x} (lo g_{10} ((x^{2} + 1)^{2}))

\int x - cos (x) d x

e^{- s} \frac{( s ^{2} + 2 )}{s ^{3}}

What is the integral of (4u)/(16u^2+7) ?
The integral of (4u)/(16u^2+7) is 1/8 ln|16u^2+7|+C