# Evaluate the integral \int ((t+1)^{2}-\frac{1}{t^{4}})dt

Evaluate the integral $$\displaystyle\int{\left({\left({t}+{1}\right)}^{{{2}}}-{\frac{{{1}}}{{{t}^{{{4}}}}}}\right)}{\left.{d}{t}\right.}$$

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

Vaing1990
Step 1
To evaluate the integral: $$\displaystyle\int{\left({\left({t}+{1}\right)}^{{{2}}}-{\frac{{{1}}}{{{t}^{{{4}}}}}}\right)}{\left.{d}{t}\right.}$$
Evaluating the given integral.
$$\displaystyle\int{\left({\left({t}+{1}\right)}^{{{2}}}-{\frac{{{1}}}{{{t}^{{{4}}}}}}\right)}{\left.{d}{t}\right.}=\int{\left[{\left({t}^{{{2}}}+{2}{t}+{1}\right)}-{\frac{{{1}}}{{{t}^{{{4}}}}}}\right]}{\left.{d}{t}\right.}$$
$$\displaystyle=\int{t}^{{{2}}}{\left.{d}{t}\right.}+{2}\int{t}{\left.{d}{t}\right.}+\int{\left.{d}{t}\right.}-\int{\frac{{{1}}}{{{t}^{{{4}}}}}}{\left.{d}{t}\right.}$$
$$\displaystyle={\frac{{{t}^{{{3}}}}}{{{3}}}}+{2}\cdot{\frac{{{t}^{{{2}}}}}{{{2}}}}+{t}+{3}{\frac{{{1}}}{{{t}^{{{3}}}}}}+{C}$$
$$\displaystyle={\frac{{{t}^{{{3}}}}}{{{3}}}}+{t}^{{{2}}}+{t}+{\frac{{{3}}}{{{t}^{{{3}}}}}}+{C}$$
Step 2
Hence, required answer is $$\displaystyle{\left[{\frac{{{t}^{{{3}}}}}{{{3}}}}+{t}^{{{2}}}+{t}+{\frac{{{3}}}{{{t}^{{{3}}}}}}\right]}+{C}$$
###### Have a similar question?
Gloria Lusk
Step 1: Expand.
$$\displaystyle\int{\left({t}+{1}\right)}^{{{2}}}-{\frac{{{1}}}{{{t}^{{{4}}}}}}{\left.{d}{t}\right.}$$
Step 2: Use Sum Rule: $$\displaystyle\int{f{{\left({x}\right)}}}+{g{{\left({x}\right)}}}{\left.{d}{x}\right.}=\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}+\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$$.
$$\displaystyle\int{\left({t}+{1}\right)}^{{{2}}}{\left.{d}{t}\right.}-\int{\frac{{{1}}}{{{t}^{{{4}}}}}}{\left.{d}{t}\right.}$$
Step 3: Expand.
$$\displaystyle\int{t}^{{{2}}}+{2}{t}+{1}{\left.{d}{t}\right.}-\int{\frac{{{1}}}{{{t}^{{{4}}}}}}{\left.{d}{t}\right.}$$
Step 4: Use Power Rule: $$\displaystyle\int{x}^{{{n}}}{\left.{d}{x}\right.}={\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}}+{C}$$.
$$\displaystyle{\frac{{{t}^{{{3}}}}}{{{3}}}}+{t}^{{{2}}}+{t}-\int{\frac{{{1}}}{{{t}^{{{4}}}}}}{\left.{d}{t}\right.}$$
Step 5: Use Power Rule: $$\displaystyle\int{x}^{{{n}}}{\left.{d}{x}\right.}={\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}}+{C}$$.
$$\displaystyle{\frac{{{t}^{{{3}}}}}{{{3}}}}+{t}^{{{2}}}+{t}+{\frac{{{1}}}{{{3}{t}^{{{3}}}}}}$$
$$\displaystyle{\frac{{{t}^{{{3}}}}}{{{3}}}}+{t}^{{{2}}}+{t}+{\frac{{{1}}}{{{3}{t}^{{{3}}}}}}+{C}$$