# Find the following indefinite integrals. (a) Integral of 3x^2 - 2/3 sqrt(x) with respect to x. ...

## Question:

Find the following indefinite integrals.

(a) {eq}\displaystyle \int \left( 3x^2 - \frac{2}{3} \sqrt{x} \right)dx {/eq}

(b) {eq}\displaystyle \int \sin(4t - 5) dt {/eq}

(c) {eq}\displaystyle \int x(x^2 + 1)^4 dx {/eq}

## Solving the following indefinite integrals:

In this problem we are asked to find the indefinite integrals.

To solve this problem we have to use some results and relations .

The results which we are going to use are {eq}\int { x^n dx } = \frac { x^{n+1}}{n+1} + {C_1} \\ \int sin x dx = - cos x + {C_2} {/eq} here C_1 , C_2 are integration constants.

a)

{eq}\begin{align*} \int \left ( 3x^2 - \frac{2}{3} \sqrt{x} \right)dx &= \int 3x^2 \; dx - \int \frac{2}{3} \sqrt{x} dx \\ &= x^3 - \frac { 2}{3 } \cdot \frac { x^{\frac { 1}{2 }+1}}{\frac { 1}{2 }+1 } + C & \text { [ Since } \int { x^n dx } = \frac { x^{n+1}}{n+1} + C \text { ]}\\ &= x^3 - \frac {4 }{9 } \cdot x^ {\frac { 3}{2 }} + C \end{align*} {/eq}

b)

{eq}\begin{align*} \int \sin(4t - 5) dt &= \frac {1 }{4 } \cdot \int 4 \sin ( 4t -5 ) dt \\ &= \frac {1 }{4 } \cdot \int \sin ( 4t -5 ) d(4t - 5) \\ &= \frac { 1}{ 4} ( - \cos (4t-5) ) + C & \text { [ Since } \int sin x dx = - cos x + C \text { ]}\\ \end{align*} {/eq}

c)

{eq}\begin{align*} \int x(x^2 + 1)^4 dx &= \frac { 1}{ 2} \cdot \int 2x(x^2 + 1)^4 dx \\ &= \frac { 1}{ 2} \cdot \int (x^2 + 1)^4 d(x^2 + 1) \\ &= \frac { 1}{ 2} \cdot \frac { (x^2+1)^5}{ 5} + C & \text { [ Since } \int { x^n dx } = \frac { x^{n+1}}{n+1} + C \text { ]}\\ &=\frac { (x^2+1)^5}{ 10} + C \end{align*} {/eq}