Solved: "Evaluate the integral ∫((t+1)2−1t4)dt"

katelineliseua

Answered question

2021-11-22

Evaluate the integral

\int ({(t + 1)}^{2} - \frac{1}{t^{4}}) d t

Vaing1990

Beginner2021-11-23Added 16 answers

Step 1
To evaluate the integral:

\int ({(t + 1)}^{2} - \frac{1}{t^{4}}) d t

Evaluating the given integral.

\int ({(t + 1)}^{2} - \frac{1}{t^{4}}) d t = \int [(t^{2} + 2 t + 1) - \frac{1}{t^{4}}] d t

= \int t^{2} d t + 2 \int t d t + \int d t - \int \frac{1}{t^{4}} d t

= \frac{t^{3}}{3} + 2 \cdot \frac{t^{2}}{2} + t + 3 \frac{1}{t^{3}} + C

= \frac{t^{3}}{3} + t^{2} + t + \frac{3}{t^{3}} + C

Step 2
Hence, required answer is

[\frac{t^{3}}{3} + t^{2} + t + \frac{3}{t^{3}}] + C

Gloria Lusk

Beginner2021-11-24Added 18 answers

Step 1: Expand.

\int {(t + 1)}^{2} - \frac{1}{t^{4}} d t

Step 2: Use Sum Rule:

\int f (x) + g (x) d x = \int f (x) d x + \int g (x) d x

\int {(t + 1)}^{2} d t - \int \frac{1}{t^{4}} d t

Step 3: Expand.

\int t^{2} + 2 t + 1 d t - \int \frac{1}{t^{4}} d t

Step 4: Use Power Rule:

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

\frac{t^{3}}{3} + t^{2} + t - \int \frac{1}{t^{4}} d t

Step 5: Use Power Rule:

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

\frac{t^{3}}{3} + t^{2} + t + \frac{1}{3 t^{3}}

Step 6: Add constant.

\frac{t^{3}}{3} + t^{2} + t + \frac{1}{3 t^{3}} + C

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