What is the integral from 1 to x^3 of (g(t))/(4+t^2) ?

The integral from 1 to x^3 of (g(t))/(4+t^2) is (g(ln|4+x^6|-ln(5)))/2

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integral from 1 to x^3 of (g(t))/(4+t^2)

\int_{1}^{x^{3}} \frac{g ( t )}{4 + t ^{2}} d t

\frac{g ( ln 4 + x ^{6} - ln ( 5 ) )}{2}

Solution steps

\int_{1}^{x^{3}} \frac{g t}{4 + t ^{2}} d t

Take the constant out:

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

= g \cdot \int_{1}^{x^{3}} \frac{t}{4 + t ^{2}} d t

Apply u-substitution

: \int_{5}^{4 + x^{6}} \frac{1}{2 u} d u

= g \cdot \int_{5}^{4 + x^{6}} \frac{1}{2 u} d u

Take the constant out:

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

= g \frac{1}{2} \cdot \int_{5}^{4 + x^{6}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= g \frac{1}{2} [ln ∣ u ∣]_{5}^{4 + x^{6}}

Compute the boundaries:

ln 4 + x^{6} - ln (5)

= g \frac{1}{2} (ln 4 + x^{6} - ln (5))

Simplify

= \frac{g ( ln 4 + x ^{6} - ln ( 5 ) )}{2}

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What is the integral from 1 to x^3 of (g(t))/(4+t^2) ?
The integral from 1 to x^3 of (g(t))/(4+t^2) is (g(ln|4+x^6|-ln(5)))/2