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)1. \(\frac{1}{2}\text{lo}{{\text{g}}_{\text{e}}}\frac{\left| {{\text{x}}^{3}}+1 \right|}{{{\text{x}}^{2}}}+\text{C}\)2. \(\frac{1}{2}\text{lo}{{\text{g}}_{\text{e}}}\frac{{{\left( {{\text{x}}^{3}}+1 \right)}^{2}}}{\left| {{\text{x}}^{3}} \right|}+\text{C}\)3. \(\text{lo}{{\text{g}}_{\text{e}}}\left| \frac{{{\text{x}}^{3}}+1}{\text{x}} \right|+\text{C}\)4. \(\text{lo}{{\text{g}}_{\text{e}}}\frac{\left| {{\text{x}}^{3}}+1 \right|}{{{\text{x}}^{2}}}+\text{C}\)

Answer» Correct Answer - Option 3 : \(\text{lo}{{\text{g}}_{\text{e}}}\left| \frac{{{\text{x}}^{3}}+1}{\text{x}} \right|+\text{C}\)

The given equation is:

\(\text{I}=\int \left( \frac{2{{\text{x}}^{3}}-1}{{{\text{x}}^{4}}+\text{x}} \right)\text{dx}\)

Now, dividing x2 in numerator and denominator,

\(\Rightarrow \text{I}=\int \left( \frac{\left( \frac{2{{\text{x}}^{3}}-1}{{{\text{x}}^{2}}} \right)}{\left( \frac{{{\text{x}}^{4}}+\text{x}}{{{\text{x}}^{2}}} \right)} \right)\text{dx}\)

\(\Rightarrow \text{I}=\int \left( \frac{2\text{x}-{{\text{x}}^{-2}}}{{{\text{x}}^{2}}+{{\text{x}}^{-1}}} \right)\text{dx}\)

\(\left[ \because \int \left( \frac{\text{{f}'}\left( \text{x} \right)}{\text{f}\left( \text{x} \right)} \right)\text{dx}={{\log }_{\text{e}}}\left( \text{f}\left( \text{x} \right) \right)+\text{C} \right]\)

⇒ I = loge(x2 + x-1) + C

\(\Rightarrow \text{I}={{\log }_{\text{e}}}\left( {{\text{x}}^{2}}+\frac{1}{\text{x}} \right)+\text{C}\)

\(\therefore \text{I}={{\log }_{\text{e}}}\left( \frac{{{\text{x}}^{3}}+1}{\text{x}} \right)+\text{C}\)


Discussion

No Comment Found

Related InterviewSolutions