The integral $\int \left( \frac{2{{\text{x}}^{3}}-1}{{{\text{x}}^{4}}+\text{x}} \right)\text{dx}$ is equal to:(Here C is a constant of integration)

Question

The integral $\int \left( \frac{2{{\text{x}}^{3}}-1}{{{\text{x}}^{4}}+\text{x}} \right)\text{dx}$ is equal to:(Here C is a constant of integration)

TuteeHUB · Accepted Answer

$\text{lo}{{\text{g}}_{\text{e}}}\frac{\left| {{\text{x}}^{3}}+1 \right|}{{{\text{x}}^{2}}}+\text{C}$

TuteeHUB · Answer

$\frac{1}{2}\text{lo}{{\text{g}}_{\text{e}}}\frac{\left| {{\text{x}}^{3}}+1 \right|}{{{\text{x}}^{2}}}+\text{C}$

TuteeHUB · Answer

$\frac{1}{2}\text{lo}{{\text{g}}_{\text{e}}}\frac{{{\left( {{\text{x}}^{3}}+1 \right)}^{2}}}{\left| {{\text{x}}^{3}} \right|}+\text{C}$

TuteeHUB · Answer

$\text{lo}{{\text{g}}_{\text{e}}}\left| \frac{{{\text{x}}^{3}}+1}{\text{x}} \right|+\text{C}$

1.	The integral \(\int \left( \frac{2{{\text{x}}^{3}}-1}{{{\text{x}}^{4}}+\text{x}} \right)\text{dx}\) is equal to:(Here C is a constant of integration)
A.	\(\frac{1}{2}\text{lo}{{\text{g}}_{\text{e}}}\frac{\left\| {{\text{x}}^{3}}+1 \right\|}{{{\text{x}}^{2}}}+\text{C}\)
B.	\(\frac{1}{2}\text{lo}{{\text{g}}_{\text{e}}}\frac{{{\left( {{\text{x}}^{3}}+1 \right)}^{2}}}{\left\| {{\text{x}}^{3}} \right\|}+\text{C}\)
C.	\(\text{lo}{{\text{g}}_{\text{e}}}\left\| \frac{{{\text{x}}^{3}}+1}{\text{x}} \right\|+\text{C}\)
D.	\(\text{lo}{{\text{g}}_{\text{e}}}\frac{\left\| {{\text{x}}^{3}}+1 \right\|}{{{\text{x}}^{2}}}+\text{C}\)
Answer» D. \(\text{lo}{{\text{g}}_{\text{e}}}\frac{\left\| {{\text{x}}^{3}}+1 \right\|}{{{\text{x}}^{2}}}+\text{C}\)

The integral \(\int \left( \frac{2{{\text{x}}^{3}}-1}{{{\text{x}}^{4}}+\text{x}} \right)\text{dx}\) is equal to:(Here C is a constant of integration)