Indefinite Integrals of Logarithmic Functions 2

Question 1 of 5

Find the integral

\int \frac{3}{2 x - 6} d x

$\int{\frac{3}{2x-6}} dx$

1. $\frac{3}{2} ln (2 x) + c$ $3/2 ln (2x)+c$
2. $3 ln (2 x - 6) + c$ $3 ln (2x-6)+c$
3. $2 ln (2 x - 6) + c$ $2 ln (2x-6)+c$
4. $\frac{3}{2} ln (2 x - 6) + c$ $3/2 ln (2x-6)+c$

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Indefinite Integral

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

First, form a fraction to balance the equation.

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$	$=$	$\int \frac{\color{#9a00c7}{3}}{\color{#D800AD}{2x-6}}$

	$=$	$3/2$	Differentiate the denominator

Use

$3/2$ as a constant to balance the integral.

$f(x)$	$=$	$3x-6$
$f'(x)$	$=$	$3$

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$	$=$	$\log_e [\color{#D800AD}{f(x)}]+c$

$\int \frac{\color{#9a00c7}{3}}{\color{#D800AD}{2x-6}}dx$	$=$	$\frac{3}{2} \ln [\color{#D800AD}{2x-6}]+c$	Substitute known values

$3/2 ln (2x-6)+c$

Question 2 of 5

Find the integral

$\int{\frac{x^2}{2x^3+4}} dx$

1. $1/3 ln (2x^3+4)+c$
2. $ln (2x^3+4)+c$
3. $1/2 ln (2x^3+4)+c$
4. $1/6 ln (2x^3+4)+c$

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Indefinite Integral

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

First, form a fraction to balance the equation.

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$	$=$	$\int \frac{\color{#9a00c7}{x^2}}{\color{#D800AD}{2x^3+4}}$

	$=$	$(x^2)/(6x^2)$	Differentiate the denominator

	$=$	$1/6$	$(x^2)/(x^2)=1$

Use

$1/6$ as a constant to balance the integral.

$f(x)$	$=$	$2x^3+4$
$f'(x)$	$=$	$6x^2$

$\frac{1}{6}\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$	$=$	$\frac{1}{6}\log_e [\color{#D800AD}{f(x)}]+c$

$\frac{1}{6}\int \frac{\color{#9a00c7}{6x^2}}{\color{#D800AD}{2x^3+4}}dx$	$=$	$\frac{1}{6} \ln [\color{#D800AD}{2x^3+4}]+c$	Substitute known values

$1/6 ln (2x^3+4)+c$

Question 3 of 5

Find the integral

$\int{\frac{x^5}{2x^6+3}} dx$

1. $ln (2x^6+3)+c$
2. $12 ln (2x^6+3)+c$
3. $1/6 ln (2x^6+3)+c$
4. $1/12 ln (2x^6+3)+c$

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Indefinite Integral

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

First, form a fraction to balance the equation.

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$	$=$	$\int \frac{\color{#9a00c7}{x^5}}{\color{#D800AD}{2x^6+3}}$

	$=$	$(x^5)/(12x^5)$	Differentiate the denominator

	$=$	$1/12$	$(x^5)/(x^5)=1$

Use

$1/12$ as a constant to balance the integral.

$f(x)$	$=$	$2x^6+3$
$f'(x)$	$=$	$12x^5$

$\frac{1}{12}\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$	$=$	$\frac{1}{12}\log_e [\color{#D800AD}{f(x)}]+c$

$\frac{1}{12}\int \frac{\color{#9a00c7}{12x^5}}{\color{#D800AD}{2x^6+3}}dx$	$=$	$\frac{1}{12} \ln [\color{#D800AD}{2x^6+3}]+c$	Substitute known values

$1/12 ln (2x^6+3)+c$

Question 4 of 5

Find the integral

$\int{\frac{x}{(x+3)(x-3)}} dx$

1. $1/2 ln (x+3)+c$
2. $1/2 ln (x-3)+c$
3. $1/2 ln (x^2-9)+c$
4. $2 ln (x^2-9)+c$

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Indefinite Integral

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

First, form a fraction to balance the equation.

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$	$=$	$\int \frac{\color{#9a00c7}{x}}{\color{#D800AD}{(x+3)(x-3)}}$

	$=$	$\frac{x}{\color{#CC0000}{x^2-9}}$	Expand

	$=$	$x/(2x)$	Differentiate the denominator

	$=$	$1/2$	$(x)/(x)=1$

Use

$1/2$ as a constant to balance the integral.

$f(x)$	$=$	$x^2-9$
$f'(x)$	$=$	$2x$

$\frac{1}{2}\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$	$=$	$\frac{1}{2}\log_e [\color{#D800AD}{f(x)}]+c$

$\frac{1}{2}\int \frac{\color{#9a00c7}{2x}}{\color{#D800AD}{x^2-9}}dx$	$=$	$\frac{1}{2} \ln [\color{#D800AD}{x^2-9}]+c$	Substitute known values

$1/2 ln (x^2-9)+c$

Question 5 of 5

Find the integral

$\int{\frac{x+3}{x^2+6x-1}} dx$

1. $ln (x^2+6x-1)+c$
2. $1/2 ln (x^2+6x-1)+c$
3. $1/2 ln (6x-1)+c$
4. $1/2 ln (x^2-6x-1)+c$

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Indefinite Integral

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx=\log_e [\color{#D800AD}{f(x)}]+c$

First, form a fraction to balance the equation.

$\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$	$=$	$\int \frac{\color{#9a00c7}{x+3}}{\color{#D800AD}{x^2+6x-1}}$

	$=$	$(x+3)/(2x+6)$	Differentiate the denominator

	$=$	$\frac{x+3}{\color{#CC0000}{2(x+3)}}$	Factorize

	$=$	$1/2$	$(x+3)/(x+3)=1$

Use

$1/2$ as a constant to balance the integral.

$f(x)$	$=$	$x^2+6x-1$
$f'(x)$	$=$	$2x+6$

$\frac{1}{2}\int \frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}dx$	$=$	$\frac{1}{2}\log_e [\color{#D800AD}{f(x)}]+c$

$\frac{1}{2}\int \frac{\color{#9a00c7}{2x+6}}{\color{#D800AD}{x^2+6x-1}}dx$	$=$	$\frac{1}{2} \ln [\color{#D800AD}{x^2+6x-1}]+c$	Substitute known values

$1/2 ln (x^2+6x-1)+c$

Indefinite Integrals of Logarithmic Functions 2

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